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


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

where:
 * $\ds {k + 1 \brace m + 1}$ and so on 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 $\map P n$ be the proposition:
 * $\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$

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:
 * $\ds \sum_k {k + 1 \brace m + 1} \binom r k \paren {-1}^{r - k} = {r \brace m}$

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

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