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

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


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

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

Proof
The proof proceeds by induction on $m$.

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

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

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

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

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