Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m

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


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

where:
 * $\displaystyle \left\{ {k + 1 \atop m + 1}\right\}$ etc. denotes a Stirling number of the second kind
 * $\dbinom n k$ denotes a binomial coefficient.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom n k \left({-1}\right)^{n - k} = \left\{ {n \atop m}\right\}$

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

So $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

When $m = 0$:

When $m = 1$:

When $m > 1$, from Stirling Number of the Second Kind of Number with Greater we have:
 * $\displaystyle \left\{ {2 \atop 2 + k}\right\} = 0$

Thus $P \left({1}\right)$ is seen to hold for all $m$.

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:
 * $\displaystyle \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r k \left({-1}\right)^{r - k} = \left\{ {r \atop m}\right\}$

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

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