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

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


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

where:
 * $\displaystyle \left[{n + 1 \atop k + 1}\right]$ etc. denotes an unsigned Stirling number of the first kind
 * $\dbinom k m$ 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:
 * $\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{n \atop m}\right]$

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

But when $m \ne 0$ we have that:
 * $\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- m} = 0 = \left[{0 \atop m}\right]$

and when $m = 0$ we have that:
 * $\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- 0} = 1 = \left[{0 \atop m}\right]$

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

from which it is to be shown that:
 * $\displaystyle \sum_k \left[{r + 2 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{r + 1 \atop m}\right]$

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