avewright/proofwiki-math huggingface dataset:
Theorem: Continuous Lattice is Meet-Continuous

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Then $L$ is meet-continuous.

Proof:
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 the axiom of approximation.
By definition of up-complete:
:$x^\ll$ admits a supremum.
By Lower Closure of Element is Ideal and definition of ideal:
:$\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ is directed.
By definition of up-complete:
:$\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ admits a supremum.
We will prove that
:$x^\ll \subseteq \paren {\sup \set {x \wedge d: d \in D} }^\preceq$
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:
:$a = a \wedge a \preceq x \wedge d \in \set {x \wedge y: y \in D}$
By definition of supremum:
:$\sup \set {x \wedge y: y \in D}$ is upper bound for $\set {x \wedge y: y \in D}$
By definition of upper bound:
:$x \wedge d \preceq \sup \set {x \wedge y: y \in D}$
By definition of transitivity:
:$a \preceq \sup \set {x \wedge y: y \in D}$
Thus by definition of lower closure of element:
:$a \in \paren {\sup \set {x \wedge d: d \in D} }^\preceq$
{{qed|lemma}}
By Supremum of Subset:
:$\map \sup {x^\ll} \preceq \map \sup {\paren {\sup \set {x \wedge d: d \in D} }^\preceq}$
By the axiom of approximation and Supremum of Lower Closure of Element:
:$x \preceq \sup \set {x \wedge d: d \in D}$
By definition:
:$x$ is upper bound for $\set {x \wedge d: d \in D}$
By definition of supremum:
:$\sup \set {x \wedge d: d \in D} \preceq x$
Thus by definition of antisymmetry:
:$x = \sup \set {x \wedge d: d \in D}$
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}}
Theorem: Number to Power of Zero Rising is One

:$x^{\overline 0} = 1$

Proof:
{{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
Theorem: Number to Power of Zero Falling is One

:$x^{\underline 0} = 1$

Proof:
{{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
Theorem: Product of Number by its Falling Factorial

Let $x^{\underline n}$ denote the $n$th falling factorial power of $x$.
Then:
:$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$

Proof:
{{begin-eqn}}
{{eqn | l = x x^{\underline n}
      | r = \paren {x - n + n} x^{\underline n}
      | c = 
}}
{{eqn | r = \paren {x - n} x^{\underline n} + n x^{\underline n}
      | c = 
}}
{{eqn | r = \paren {x - n} \prod_{j \mathop = 0}^{n - 1} \paren {x - j} + n x^{\underline n}
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^n \paren {x - j} + n x^{\underline n}
      | c = {{Defof|Continued Product}}
}}
{{eqn | r = x^{\underline {n + 1} } + n x^{\underline n}
      | c = {{Defof|Falling Factorial}}
}}
{{end-eqn}}
{{qed}}
Category:Falling Factorials
8jeuowxyipvgloeua0fcnalc3snbysa
Theorem: Product of Number by its Rising Factorial

Let $x^{\overline n}$ denote the $n$th rising factorial power of $x$.
Then:
:$x x^{\overline n} = x^{\overline {n + 1} } - n x^{\overline n}$

Proof:
{{begin-eqn}}
{{eqn | l = x x^{\overline n}
      | r = \paren {x + n - n} x^{\overline n}
      | c = 
}}
{{eqn | r = \paren {x + n} x^{\overline n} - n x^{\overline n}
      | c = 
}}
{{eqn | r = \paren {x + n} \prod_{j \mathop = 0}^{n - 1} \paren {x + j} - n x^{\overline n}
      | c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^n \paren {x + j} - n x^{\overline n}
      | c = {{Defof|Continued Product}}
}}
{{eqn | r = x^{\overline {n + 1} } - n x^{\overline n}
      | c = {{Defof|Rising Factorial}}
}}
{{end-eqn}}
{{qed}}
Category:Rising Factorials
jzuzu2mgx54dnae41nbbl42y3myh8nz
Theorem: First Inversion Formula for Stirling Numbers

For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\delta_{m n}$ denotes the Kronecker delta.

Proof:
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
Theorem: Unsigned Stirling Number of the First Kind of 1

:$\ds {1 \brack n} = \delta_{1 n}$

Proof:
{{begin-eqn}}
{{eqn | l = {1 \brack n}
      | r = 0 \times {0 \brack n} + {0 \brack n - 1}
      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | r = {0 \brack n - 1}
      | c = 
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | r = \delta_{1 n}
      | c = $0 = n - 1 \iff 1 = n$
}}
{{end-eqn}}
{{qed}}
Theorem: Signed Stirling Number of the First Kind of 1

:$\map s {1, n} = \delta_{1 n}$

Proof:
{{begin-eqn}}
{{eqn | l = \map s {1, n}
      | r = \map s {0, n - 1} - 0 \times \map s {0, n}
      | c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \map s {0, n - 1}
      | c =
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
      | c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \delta_{1 n}
      | c = $0 = n - 1 \iff 1 = n$
}}
{{end-eqn}}
{{qed}}
Theorem: Stirling Number of the Second Kind of 1

:$\ds {1 \brace n} = \delta_{1 n}$

Proof:
{{begin-eqn}}
{{eqn | l = {1 \brace n}
      | r = n \times {0 \brace n} + {0 \brace n - 1}
      | c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = n \times \delta_{0 n} + \delta_{0 \paren {n - 1} }
      | c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
      | c = as $n \times \delta_{0 n}$ is either $n \times 0$ or $0 \times 1$ and so always $0$
}}
{{eqn | r = \delta_{1 n}
      | c = $0 = n - 1 \iff 1 = n$
}}
{{end-eqn}}
{{qed}}
Theorem: Second Inversion Formula for Stirling Numbers

For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brace k}$ denotes a Stirling number of the second kind
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\delta_{m n}$ denotes the Kronecker delta.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
Theorem: Zero Choose n

:$\dbinom 0 n = \delta_{0 n}$

Proof:
:$\dbinom m n = \begin{cases}\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\&\\0 & : \text { otherwise } \end{cases}$
Theorem: Unsigned Stirling Number of the First Kind of 0

:$\ds {0 \brack n} = \delta_{0 n}$

Proof:
By definition of unsigned Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$
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 = {{Defof|Integer Power}}
}}
{{end-eqn}}
Thus, in the expression:
:$\ds x^{\underline 0} = \sum_k \paren {-1}^{-k} {0 \brack k} x^k$
we have:
:$\ds {0 \brack 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brack k} = 0$
That is:
:$\ds {0 \brack k} = \delta_{0 k}$
{{qed}}
Theorem: Signed Stirling Number of the First Kind of 0

:$\map s {0, n} = \delta_{0 n}$

Proof:
By definition of signed Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
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 = {{Defof|Integer Power}}
}}
{{end-eqn}}
Thus, in the expression:
:$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
we have:
:$\map s {0, 0} = 1$
and for all $k \in \Z_{>0}$:
:$\map s {0, k} = 0$
That is:
:$\map s {0, k} = \delta_{0 k}$
{{qed}}
Theorem: Stirling Number of the Second Kind of 0

:$\ds {0 \brace n} = \delta_{0 n}$

Proof:
By definition of Stirling numbers of the second kind:
$\ds x^{\underline 0} = \sum_k {0 \brace k} x^k$
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:
:$\ds x^0 = \sum_k {0 \brace k} x^{\underline k}$
we have:
:$\ds {0 \brace 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brace k} = 0$
That is:
:$\ds {0 \brace k} = \delta_{0 k}$
{{qed}}
Theorem: Unsigned Stirling Number of the First Kind of Number with Self

:$\ds {n \brack n} = 1$

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n} = 1$
Theorem: Unsigned Stirling Number of the First Kind of Number with Greater

Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
:$\ds {n \brack k} = 0$

Proof:
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
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 $\ds {n \brack k}$ of $x^k$ where $k > n$ is $0$.
{{qed}}
Theorem: Unsigned Stirling Number of the First Kind of Number with Greater

Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
:$\ds {n \brack k} = 0$

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brack k} = 0$
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:
:$\ds k > r \implies {r \brack k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brack k} = 0$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = {r + 1 \brack k}
      | r = r {r \brack k} + {r \brack k - 1}
      | c = 
}}
{{eqn | r = r \times 0 + {r \brack 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:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brack k} = 0$
{{qed}}
Theorem: Signed Stirling Number of the First Kind of Number with Greater

Let $\map s {n, k}$ denote a signed Stirling number of the first kind.
Then:
:$\map s {n, k} = 0$

Proof:
By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $\map s {n, k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \map s {n, k} x^k$
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 $\map s {n, k}$ of $x^k$ where $k > n$ is $0$.
{{qed}}
Theorem: Stirling Number of the Second Kind of Number with Greater

Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$

Proof:
By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
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 $\ds {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
{{qed}}
Theorem: Stirling Number of the Second Kind of Number with Greater

Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from Stirling Number of the Second Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brace k} = 0$
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:
:$\ds k > r \implies {r \brace k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brace k} = 0$
=== 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:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$
{{qed}}
Theorem: Stirling Number of the Second Kind of Number with Self

:$\ds {n \brace n} = 1$

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n} = 1$
Theorem: Value of Term under Assignment Determined by Variables

Let $\tau$ be a term of predicate logic.
Let $\AA$ be a structure for predicate logic.
Let $\sigma, \sigma'$ be assignments for $\tau$ in $\AA$ such that:
:For each variable $x$ occurring in $\tau$, $\map \sigma x = \map {\sigma'} x$
Then:
:$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = \map {\operatorname{val}_\AA} \tau \sqbrk {\sigma'}$
where $\map {\operatorname{val}_\AA} \tau \sqbrk \sigma$ is the value of $\tau$ under $\sigma$.

Proof:
Proceed by the Principle of Structural Induction applied to the definition of a term.
If $\tau = x$, then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma
      | r = \map \sigma x
      | c = {{Defof|Value of Term under Assignment|value under $\sigma$}}
}}
{{eqn | r = \map {\sigma'} x
      | c = Assumption on $\sigma, \sigma'$
}}
{{eqn | r = \map {\operatorname{val}_\AA} \tau \sqbrk {\sigma'}
      | c = {{Defof|Value of Term under Assignment|value under $\sigma$}}
}}
{{end-eqn}}
as desired.
If $\tau = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis applies to each $\tau_i$, then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma
      | r = \map {f_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk \sigma}
      | c = {{Defof|Value of Term under Assignment|value under $\sigma$}}
}}
{{eqn | r = \map {f_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk {\sigma'}, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk {\sigma'} }
      | c = Induction Hypothesis
}}
{{eqn | r = \map {\operatorname{val}_\AA} \tau \sqbrk {\sigma'}
      | c = {{Defof|Value of Term under Assignment|value under $\sigma'$}}
}}
{{end-eqn}}
The result follows from the Principle of Structural Induction.
{{qed}}
Category:Predicate Logic
m1rk6i4ud6bel7gcky326ia2cvl83qh
Theorem: Relation between Signed and Unsigned Stirling Numbers of the First Kind

Let $m, n \in \Z_{\ge 0}$ be positive integers.
Then:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
where:
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind
:$\map s {n, m}$ denotes a signed Stirling number of the first kind.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_k \map s {n, k} x^k
      | r = x^{\underline n}
      | c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | ll= \leadsto
      | l = \map s {n, m}
      | r = \paren {-1}^{n - m} {n \brack m}
      | c = Comparing coefficients of $x^m$
}}
{{eqn | ll= \leadsto
      | l = {n \brack m}
      | r = \paren {-1}^{m - n} \map s {n, m}
}}
{{eqn | r = \paren {-1}^{n + m} \map s {n, m}
      | c = as $\paren {-1}^{2 n} = 1$
}}
{{end-eqn}}
{{qed}}
Category:Stirling Numbers
1phz3bhvvgfb60lfwdv9u6hyncc95tk
Theorem: Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1

Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}}

Proof:
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
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 $\ds {n \brack k}$ of $x^k$ where $k > n$ is $0$.
{{qed}}
Theorem: Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2

Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}}

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brack k} = 0$
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:
:$\ds k > r \implies {r \brack k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brack k} = 0$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = {r + 1 \brack k}
      | r = r {r \brack k} + {r \brack k - 1}
      | c = 
}}
{{eqn | r = r \times 0 + {r \brack 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:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brack k} = 0$
{{qed}}
Theorem: Stirling Number of the Second Kind of Number with Greater/Proof 1

Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}}

Proof:
By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
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 $\ds {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
{{qed}}
Theorem: Stirling Number of the Second Kind of Number with Greater/Proof 2

Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}}

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from Stirling Number of the Second Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brace k} = 0$
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:
:$\ds k > r \implies {r \brace k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brace k} = 0$
=== 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:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$
{{qed}}
Theorem: Signed Stirling Number of the First Kind of Number with Self

:$\map s {n, n} = 1$

Proof:
From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n} = \paren {-1}^{n + n} \map s {n, n}$
We have that:
:$\paren {-1}^{n + n} = \paren {-1}^{2 n} = 1$
and so:
:$\ds {n \brack n} = \map s {n, n}$
The result follows from Unsigned Stirling Number of the First Kind of Number with Self.
{{qed}}
Theorem: Unsigned Stirling Number of the First Kind of n with n-1

:$\ds {n \brack n - 1} = \binom n 2$

Proof:
The proof proceeds by induction.
Theorem: Stirling Number of the Second Kind of n with n-1

:$\ds {n \brace n - 1} = \binom n 2$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 1} = \binom n 2$
Theorem: Signed Stirling Number of the First Kind of n with n-1

:$\map s {n, n - 1} = -\dbinom n 2$

Proof:
From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$
where $\ds {n \brack n - 1}$ 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:
:$\ds {n \brack n} = -\map s {n, n}$
The result follows from Unsigned Stirling Number of the First Kind of Number with Self.
{{qed}}
Theorem: Unsigned Stirling Number of the First Kind of n+1 with 0

:$\ds {n + 1 \brack 0} = 0$

Proof:
We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brack k} = \delta_{n k}$
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}}
Theorem: Stirling Number of the Second Kind of n+1 with 0

:$\ds {n + 1 \brace 0} = 0$

Proof:
We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brace k} = \delta_{n k}$
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}}