Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k

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


 * $\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$

where:
 * $\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
 * $\ds {m + k \brace k}$ denotes a Stirling number of the second kind
 * $\ds {n \brack n - m}$ denotes an unsigned Stirling number of the first kind.

Proof
The proof proceeds by induction on $m$.

For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$

$\map P 0$ is the case:

So $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold for all $m$.

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:
 * $\ds \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brace k} = {n \brack n - r}$

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

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