The proof proceeds by induction. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :the coefficients of the falling factorial powers  in the expression $\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$ are uniquely defined by $\displaystyle {n \brace k} =  {n - 1 \brace k - 1} + k {n - 1 \brace k}$ where $\displaystyle {n \brace k} = \delta_{n k}$ where $k = 0$ or $n = 0$. First the case where $n = 0$ is attended to. 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 }} {{eqn | r = \sum_k \delta_{0 k} x^{\underline k}       | c = {{Defof|Kronecker Delta}} }} {{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: :$\displaystyle {0 \brace k} = \delta_{0 k}$ Hence the result holds for $n = 0$. === Basis for the Induction === $\map P 1$ is the case: We have: {{begin-eqn}} {{eqn | l = x^1       | r = x       | c = {{Defof|Integer Power}} }} {{eqn | r = x^{\underline 1}       | c = Number to Power of One Falling is Itself }} {{eqn | r = \sum_k \delta_{1 k} x^{\underline k}       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = {1 \brace k}       | r = {0 \brace k - 1} + k {0 \brace k}       | c = by hypothesis }} {{eqn | r = \delta_{0 \paren {k - 1} } + k \delta_{0 k}       | c = by hypothesis }} {{eqn | r = \delta_{0 \paren {k - 1} }       | c = as $k \delta_{0 k} = 0$ for all $k$ }} {{eqn | r = \delta_{1 k}       | c =  }} {{end-eqn}} Thus, in the expression: :$\displaystyle x^1 = \sum_k {1 \brace k} x^{\underline k}$ we have: :$\displaystyle {1 \brace 1} = 1$ and for all $k \in \Z$ where $k \ne 1$: :$\displaystyle {1 \brace k} = 0$ That is: :$\displaystyle {1 \brace k} = \delta_{1 k}$ Thus $\map P 1$ 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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :The coefficients in the expression $\displaystyle x^r = \sum_k {r \brace k} x^{\underline k}$ are uniquely defined by $\displaystyle {r \brace k} = {r - 1 \brace k - 1} + k {r - 1 \brace k}$ from which it is to be shown that: :The coefficients in the expression $\displaystyle x^{r + 1} = \sum_k {r + 1 \brace k} x^{\underline k}$ are uniquely defined by $\displaystyle {r + 1 \brace k} = {r \brace k - 1} + k {r \brace k}$ === Induction Step === This is the induction step: Anticipating the expected result, we use $\displaystyle {r + 1 \brace k}$ to denote the coefficients of the $k$th falling factorial power in the expansion of $x^{r + 1}$. Thus: {{begin-eqn}} {{eqn | l = x^{r + 1}       | r = x x^r       | c =  }} {{eqn | r = x \left({\sum_k {r \brace k} x^{\underline k} }\right)       | c = Induction Hypothesis }} {{eqn | r = \sum_k {r \brace k} x x^{\underline k}       | c = $x$ is constant in this context }} {{eqn | r = \sum_k {r \brace k} \paren {x^{\underline {k + 1} } + k x^{\underline k} }       | c = Product of Number by its Falling Factorial }} {{eqn | r = \sum_k {r \brace k} x^{\underline {k + 1} } + \sum_k {r \brace k} k x^{\underline k}       | c = Sum of Summations equals Summation of Sum }} {{eqn | r = \sum_k {r \brace k - 1}  x^{\underline k} + \sum_k {r \brace k} k x^{\underline k}       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_k \paren { {r \brace k - 1} x^{\underline k} + {r \brace k} k x^{\underline k} }       | c = Sum of Summations equals Summation of Sum }} {{eqn | r = \sum_k \paren { {r \brace k - 1} + k {r \brace k} } x^{\underline k}       | c =  }} {{end-eqn}} Thus the coefficients of the falling factorial powers are defined by the recurrence relation: :$\displaystyle {r + 1 \brace k} = {r \brace k - 1} + k {r \brace k}$ as required. So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :for all $n \in \Z_{\ge 0}$, the coefficients of the falling factorial powers  in the expression $\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$ are uniquely defined by: ::$\displaystyle {n \brace k} =  {n - 1 \brace k - 1} + k {n - 1 \brace  k}$ :where $\displaystyle {n \brace k} = \delta_{n k}$ where $k = 0$ or $n = 0$. {{qed}}