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


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

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

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

But when $m \ne 0$ we have that:
 * $\ds {0 \brack m} \paren {-1}^{- m} = 0 = {0 \brack m}$

and when $m = 0$ we have that:
 * $\ds {0 \brack m} \paren {-1}^{- 0} = 1 = {0 \brack m}$

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

from which it is to be shown that:
 * $\ds \sum_k {r + 2 \brack k + 1} \binom k m \paren {-1}^{k - m} = {r + 1 \brack m}$

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