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

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


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

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

Proof
The proof proceeds by induction on $n$.

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

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

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

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