Sum over k of m choose k by -1^m-k by k to the n

Theorem
Let $m, n \in \Z_{\ge 0}$.


 * $\displaystyle \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$

where:
 * $\dbinom m k$ denotes a binomial coefficient
 * $\displaystyle {n \brace m}$ etc. denotes a Stirling number of the second kind
 * $m!$ denotes a factorial.

Proof
The proof proceeds by induction on $m$.

For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$

Basis for the Induction
$\map P 0$ is the case:

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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_k \binom r k \paren {-1}^{r - k} k^n = r! {n \brace r}$

from which it is to be shown that:
 * $\displaystyle \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n = \paren {r + 1}! {n \brace r + 1}$

Induction Step
This is the induction step:

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$