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

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


 * $\displaystyle \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$

where:
 * $\dbinom m k$ denotes a binomial coefficient
 * $\displaystyle \left\{ {n \atop m}\right\}$ 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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$

Basis for the Induction
$P \left({0}\right)$ is the case:

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 $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} k^n = r! \left\{ {n \atop r}\right\}$

from which it is to be shown that:
 * $\displaystyle \sum_k \binom {r + 1} k \left({-1}\right)^{r + 1 - k} k^n = \left({r + 1}\right)! \left\{ {n \atop r + 1}\right\}$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$