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