hcju/proofwikiqa huggingface dataset:
Let $x \in S$, $D$ be a directed subset of $S$ such that :$x \preceq \sup D$ By Way Below Closure is Directed in Bounded Below Join Semilattice: :$x^\ll$ is directed. By definition of continuous: :$L$ is up-complete and satisfies axiom of approximation. By definition of up-complete: :$x^\ll$ admits a supremum. By Lower Closure of Element is Ideal and definition of ideal: :$\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ is directed. By definition of up-complete: :$\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ admits a supremum. We will prove that :$x^\ll \subseteq \left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ Let $a \in x^\ll$ By definition of way below closure: :$a \ll x$ By Way Below implies Preceding: :$a \preceq x$ By definition of way below relation: :$\exists d \in D: a \preceq d$ By Meet is Idempotent and Meet Semilattice is Ordered Structure: :$a = a \wedge a \preceq x \wedge d \in \left\{ {x \wedge y: y \in D}\right\}$ By definition of supremum: :$\sup \left\{ {x \wedge y: y \in D}\right\}$ is upper bound for $\left\{ {x \wedge y: y \in D}\right\}$ By definition of upper bound: :$x \wedge d \preceq \sup \left\{ {x \wedge y: y \in D}\right\}$ By definition of transitivity :$a \preceq \sup \left\{ {x \wedge y: y \in D}\right\}$ Thus by definition of lower closure of element: :$a \in \left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ {{qed|lemma}} By Supremum of Subset: :$\sup \left({x^\ll}\right) \preceq \sup \left({\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq}\right)$ By axiom of approximation and Supremum of Lower Closure of Element: :$x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$ By definition: :$x$ is upper bound for $\left\{ {x \wedge d: d \in D}\right\}$ By definition of supremum: :$\sup \left\{ {x \wedge d: d \in D}\right\} \preceq x$ Thus by definition of antisymmetry: :$x = \sup \left\{ {x \wedge d: d \in D}\right\}$ Thus by Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset: :$L$ is meet-continuous. {{qed}}
{{begin-eqn}} {{eqn | l = x^{\overline 0}       | r = \prod_{j \mathop = 0}^{-1} \paren {x + j}       | c = {{Defof|Rising Factorial}} }} {{eqn | r = 1       | c = Product is Vacuous }} {{end-eqn}} {{qed}} Category:Rising Factorials 1tfwsij49sdapn8zmyovlleahasydr4
{{begin-eqn}} {{eqn | l = x^{\underline 0}       | r = \prod_{j \mathop = 0}^{-1} \paren {x - j}       | c = {{Defof|Falling Factorial}} }} {{eqn | r = 1       | c = Product is Vacuous }} {{end-eqn}} {{qed}} Category:Falling Factorials 0thb0bl3pugwn0hxx3fa89xkjtu0dbp
{{begin-eqn}} {{eqn | l = x x^{\underline n}       | r = \left({x - n + n}\right) x^{\underline n}       | c =  }} {{eqn | r = \left({x - n}\right) x^{\underline n} + n x^{\underline n}       | c =  }} {{eqn | r = \left({x - n}\right) \prod_{j \mathop = 0}^{n - 1} \left({x - j}\right) + n x^{\underline n}       | c = Definition of Falling Factorial }} {{eqn | r = \prod_{j \mathop = 0}^n \left({x - j}\right) + n x^{\underline n}       | c = Definition of Product }} {{eqn | r = x^{\underline {n + 1} } + n x^{\underline n}       | c = Definition of Falling Factorial }} {{end-eqn}} {{qed}} Category:Falling Factorials 9vaivzg0rwkuq3jc54wr842zac0fg21
{{begin-eqn}} {{eqn | l = s \left({1, n}\right)       | r = s \left({0, {n - 1} }\right) - 0 \times s \left({0, n}\right)       | c = Definition of Signed Stirling Number of the First Kind }} {{eqn | r = s \left({0, {n - 1} }\right)       | c =  }} {{eqn | r = \delta_{0 \left({n - 1}\right)}       | c = Definition of Signed Stirling Number of the First Kind }} {{eqn | r = \delta_{1 n}       | c = $0 = n - 1 \iff 1 = n$ }} {{end-eqn}} {{qed}}
By definition of binomial coefficient: :$\dbinom m n = \begin{cases} \dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\ & \\ 0 & : \text { otherwise } \end{cases}$ Thus when $n > 0$: :$\dbinom 0 n = 0$ and when $n = 0$: :$\dbinom 0 0 = \dfrac {0!} {0! \paren {0 - 0}!} = 1$ by definition of factorial. Hence the result by definition of the Kronecker delta. {{qed}}
By definition of unsigned Stirling number of the first kind: $\displaystyle x^{\underline 0} = \sum_k \left({-1}\right)^{0 - k} \left[{0 \atop k}\right] x^k$ Thus we have: {{begin-eqn}} {{eqn | l = x^{\underline 0}       | r = 1       | c = Number to Power of Zero Falling is One }} {{eqn | r = x^0       | c = Definition of Integer Power }} {{end-eqn}} Thus, in the expression: :$\displaystyle x^{\underline 0} = \sum_k \left({-1}\right)^{-k} \left[{0 \atop k}\right] x^k$ we have: :$\displaystyle \left[{0 \atop 0}\right] = 1$ and for all $k \in \Z_{>0}$: :$\displaystyle \left[{0 \atop k}\right] = 0$ That is: :$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$ {{qed}}
By definition of signed Stirling number of the first kind: $\displaystyle x^{\underline 0} = \sum_k s \left({0, k}\right) x^k$ Thus we have: {{begin-eqn}} {{eqn | l = x^{\underline 0}       | r = 1       | c = Number to Power of Zero Falling is One }} {{eqn | r = x^0       | c = Definition of Integer Power }} {{end-eqn}} Thus, in the expression: :$\displaystyle x^{\underline 0} = \sum_k s \left({0, k}\right) x^k$ we have: :$s \left({0, 0}\right) = 1$ and for all $k \in \Z_{>0}$: :$s \left({0, k}\right) = 0$ That is: :$s \left({0, k}\right) = \delta_{0 k}$ {{qed}}
By definition of Stirling numbers of the second kind: $\displaystyle x^{\underline 0} = \sum_k {0 \brace k} x^k$ Thus we have: {{begin-eqn}} {{eqn | l = x^0       | r = 1       | c = {{Defof|Integer Power}} }} {{eqn | r = x^{\underline 0}       | c = Number to Power of Zero Falling is One }} {{end-eqn}} Thus, in the expression: :$\displaystyle x^0 = \sum_k {0 \brace k} x^{\underline k}$ we have: :$\displaystyle {0 \brace 0} = 1$ and for all $k \in \Z_{>0}$: :$\displaystyle {0 \brace k} = 0$ That is: :$\displaystyle {0 \brace k} = \delta_{0 k}$ {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n \atop n}\right] = 1$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left[{0 \atop 0}\right]       | r = \delta_{0 0}       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = 1       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{k \atop k}\right] = 1$ from which it is to be shown that: :$\displaystyle \left[{ {k + 1} \atop {k + 1} }\right] = 1$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{ {k + 1} \atop {k + 1} }\right]       | r = k \left[{k \atop {k + 1} }\right] + \left[{k \atop k}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = k \times 0 + \left[{k \atop k}\right]       | c = Stirling Number of Number with Greater }} {{eqn | r = \left[{k \atop k}\right]       | c =  }} {{eqn | r = 1       | c = Induction Hypothesis }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \left[{n \atop n}\right] = 1$ {{qed}}
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\displaystyle \left[{n \atop k}\right]$ which satisfy the equation: :$\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$ where $x^{\underline n}$ denotes the $n$th falling factorial of $x$. Both of the expressions on the {{LHS}} and {{RHS}} are polynomials in $x$ of degree $n$. Hence the coefficient $\displaystyle \left[{n \atop k}\right]$ of $x^k$ where $k > n$ is $0$. {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle k > n \implies \left[{n \atop k}\right] = 0$ === Basis for the Induction === $P \left({0}\right)$ is the case: :$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$ from Unsigned Stirling Number of the First Kind of 0. So by definition of Kronecker delta: :$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle \left[{0 \atop k}\right] = 0$ and so $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $0 \le r$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle k > r \implies \left[{r \atop k}\right] = 0$ from which it is to be shown that: :$\displaystyle k > r + 1 \implies \left[{r + 1 \atop k}\right] = 0$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{r + 1 \atop k}\right]       | r = r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]       | c =  }} {{eqn | r = r \times 0 + \left[{r \atop k - 1}\right]       | c = Induction Hypothesis }} {{eqn | r = r \times 0 + 0       | c = Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies \left[{n \atop k}\right] = 0$ {{qed}}
By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $\map s {n, k}$ which satisfy the equation: :$\displaystyle x^{\underline n} = \sum_k \map s {n, k} x^k$ where $x^{\underline n}$ denotes the $n$th falling factorial of $x$. Both of the expressions on the {{LHS}} and {{RHS}} are polynomials in $x$ of degree $n$. Hence the coefficient $s \left({n, k}\right)$ of $x^k$ where $k > n$ is $0$. {{qed}}
By definition, the Stirling numbers of the second kind are defined as the coefficients $\displaystyle {n \brace k}$ which satisfy the equation: :$\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$ where $x^{\underline k}$ denotes the $k$th falling factorial of $x$. Both of the expressions on the {{LHS}} and {{RHS}} are polynomials in $x$ of degree $n$. Hence the coefficient $\displaystyle {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$. {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle k > n \implies {n \brace k} = 0$ === Basis for the Induction === $\map P 0$ is the case: :$\displaystyle {0 \brace k} = \delta_{0 k}$ from Stirling Number of the Second Kind of 0. So by definition of Kronecker delta: :$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle {0 \brace k} = 0$ and so $\map P 0$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle k > r \implies {r \brace k} = 0$ from which it is to be shown that: :$\displaystyle k > r + 1 \implies {r + 1 \brace k} = 0$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {r + 1 \brace k}       | r = k {r \brace k} + {r \brace k - 1}       | c =  }} {{eqn | r = r \times 0 + {r \brace k - 1}       | c = Induction Hypothesis }} {{eqn | r = r \times 0 + 0       | c = Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$ {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle {n \brace n} = 1$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = {0 \brace 0}       | r = \delta_{0 0}       | c = Stirling Number of the Second Kind of 0 }} {{eqn | r = 1       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k \brace k} = 1$ from which it is to be shown that: :$\displaystyle {k + 1 \brace k + 1} = 1$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 1 \brace k + 1}       | r = \paren {k + 1} {k \brace k + 1} + {k \brace k}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = \paren {k + 1} \times 0 + {k \brace k}       | c = Stirling Number of Number with Greater }} {{eqn | r = {k \brace k}       | c =  }} {{eqn | r = 1       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: {n \brace n} = 1$ {{qed}}
Proceed by the Principle of Structural Induction applied to the definition of a term. If $\tau = x$, then: {{begin-eqn}} {{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]      |r = \sigma \left({x}\right)      |c = Definition of value under $\sigma$ }} {{eqn|r = \sigma' \left({x}\right)      |c = Assumption on $\sigma, \sigma'$ }} {{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]      |c = Definition of value under $\sigma$ }} {{end-eqn}} as desired. If $\tau = f \left({\tau_1, \ldots, \tau_n}\right)$ and the induction hypothesis applies to each $\tau_i$, then: {{begin-eqn}} {{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]      |r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma}\right] }\right)      |c = Definition of value under $\sigma$ }} {{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma'}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma'}\right] }\right)      |c = Induction Hypothesis }} {{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]      |c = Definition of value under $\sigma'$ }} {{end-eqn}} The result follows from the Principle of Structural Induction. {{qed}} Category:Predicate Logic 10ma1jy4vmkw964isbnsakdzd6mfpgc
From Relation between Signed and Unsigned Stirling Numbers of the First Kind: :$\displaystyle \left[{n \atop n}\right] = \left({-1}\right)^{n + n} s \left({n, n}\right)$ We have that: :$\left({-1}\right)^{n + n} = \left({-1}\right)^{2 n} = 1$ and so: :$\displaystyle \left[{n \atop n}\right] = s \left({n, n}\right)$ The result follows from Unsigned Stirling Number of the First Kind of Number with Self. {{qed}}
The proof proceeds by induction. === Basis for the Induction === For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n \atop n - 1}\right] = \binom n 2$ $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left[{1 \atop 0}\right]       | r = \delta_{1 0}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = \binom 1 2       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{k \atop k - 1}\right] = \binom k 2$ from which it is to be shown that: :$\displaystyle \left[{ {k + 1} \atop k}\right] = \binom {k + 1} 2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{ {k + 1} \atop k}\right]       | r = k \left[{k \atop k}\right] + \left[{k \atop k - 1}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = k \times 1 + \left[{k \atop k - 1}\right]       | c = Unsigned Stirling Number of the First Kind of Number with Self }} {{eqn | r = k + \binom k 2       | c = Induction Hypothesis }} {{eqn | r = k + \frac {k \left({k - 1}\right)} 2       | c = Binomial Coefficient with Two }} {{eqn | r = \frac {k \left({k - 1}\right) + 2 k} 2       | c =  }} {{eqn | r = \frac {k^2 - k + 2 k} 2       | c =  }} {{eqn | r = \frac {\left({k + 1}\right) k} 2       | c =  }} {{eqn | r = \binom {k + 1} 2       | c = Binomial Coefficient with Two }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 1}: \left[{n \atop n - 1}\right] = \binom n 2$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\displaystyle {n \brace n - 1} = \binom n 2$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = {1 \brace 0}       | r = \delta_{1 0}       | c = Stirling Number of the Second Kind of 1 }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = \binom 1 2       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k \brace k - 1} = \binom k 2$ from which it is to be shown that: :$\displaystyle {k + 1 \brace k} = \binom {k + 1} 2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 1 \brace k}       | r = k {k \brace k} + {k \brace k - 1}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = k \times 1 + {k \brace k - 1}       | c = Stirling Number of the Second Kind of Number with Self }} {{eqn | r = k + \binom k 2       | c = Induction Hypothesis }} {{eqn | r = k + \frac {k \paren {k - 1} } 2       | c = Binomial Coefficient with Two }} {{eqn | r = \frac {k \paren {k - 1} + 2 k} 2       | c =  }} {{eqn | r = \frac {k^2 - k + 2 k} 2       | c =  }} {{eqn | r = \frac {\paren {k + 1} k} 2       | c =  }} {{eqn | r = \binom {k + 1} 2       | c = Binomial Coefficient with Two }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 1}: {n \brace n - 1} = \binom n 2$ {{qed}}
From Relation between Signed and Unsigned Stirling Numbers of the First Kind: :$\displaystyle {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$ where $\displaystyle {n \brack n - 1}$ denotes an unsigned Stirling number of the first kind. We have that: :$\paren {-1}^{n + n - 1} = \paren {-1}^{2 n - 1} = -1$ and so: :$\displaystyle {n \brack n} = -\map s {n, n}$ The result follows from Unsigned Stirling Number of the First Kind of Number with Self. {{qed}}
We are given that $k = 0$. So by definition of unsigned Stirling number of the first kind: :$\displaystyle {n \brack k} = \delta_{n k}$ where $\delta_{n k}$ is the Kronecker delta. Thus: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 0       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = n + 1       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = n + 1       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \delta_{\paren {n + 1} 0}       | r = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}}
We are given that $k = 0$. So by definition of unsigned Stirling number of the first kind: :$\displaystyle {n \brace k} = \delta_{n k}$ where $\delta_{n k}$ is the Kronecker delta. Thus {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 0       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = n + 1       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = n + 1       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \delta_{\paren {n + 1} 0}       | r = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}}
By definition of signed Stirling number of the first kind: :$\map s {n, k} = \delta_{n k}$ where $\delta_{n k}$ is the Kronecker delta. Thus {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 0       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = n + 1       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = n + 1       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \delta_{\paren {n + 1} 0}       | r = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n + 1 \atop 1}\right] = n!$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left[{1 \atop 1}\right]       | r = \delta_{1 1}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = 1       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = 0!       | c = {{Defof|Factorial}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{k + 1 \atop 1}\right] = k!$ from which it is to be shown that: :$\displaystyle \left[{k + 2 \atop 1}\right] = \left({k + 1}\right)!$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{k + 2 \atop 1}\right]       | r = \left({k + 1}\right) \left[{k + 1 \atop 1}\right] + \left[{k + 1 \atop 0}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = \left({k + 1}\right) \left[{k + 1 \atop 1}\right] + 0       | c = Unsigned Stirling Number of the First Kind of n+1 with 0 }} {{eqn | r = \left({k + 1}\right) \times k!       | c = Induction Hypothesis }} {{eqn | r = \left({k + 1}\right)!       | c = {{Defof|Factorial}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \left[{n + 1 \atop 1}\right] = n!$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle {n + 1 \brace 1} = 1$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = {1 \brace 1}       | r = \delta_{1 1}       | c = Stirling Number of the Second Kind of 1 }} {{eqn | r = 1       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k + 1 \brace 1} = 1$ from which it is to be shown that: :$\displaystyle {k + 2 \brace 1} = 1$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 2 \brace 1}       | r = 1 \times {k + 1 \brace 1} + {k + 2 \brace 0}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = {k + 1 \brace 1} + 0       | c = Stirling Number of the Second Kind of n+1 with 0 }} {{eqn | r = 1       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$ {{qed}}
By Relation between Signed and Unsigned Stirling Numbers of the First Kind: :$\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^{n + 1 + 1} s \left({n + 1, 1}\right)$ where $\displaystyle \left[{n + 1 \atop 1}\right]$ denotes an unsigned Stirling number of the first kind. We have that: :$\left({-1}\right)^{n + 1 + 1} = \left({-1}\right)^n$ and so: :$\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^n s \left({n + 1, 1}\right)$ The result follows from Unsigned Stirling Number of the First Kind of Number with Self. {{qed}}