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


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

where:
 * $\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
 * $\displaystyle \left\{ {m + k \atop k}\right\}$ denotes a Stirling number of the second kind
 * $\displaystyle \left[{n \atop n - m}\right]$ 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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} \left\{ {m + k \atop k}\right\} = \left[{n \atop n - m}\right]$

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

So $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold for all $m$.

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:
 * $\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} \left\{ {r + k \atop k}\right\} = \left[{n \atop n - r}\right]$

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

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