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

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


 * $\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$

where $\ds {k \brace m}$ etc. denotes a Stirling number of the second kind.

Proof
The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$

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:
 * $\ds \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {r + 1 \brace m + 1}$

from which it is to be shown that:
 * $\ds \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = {r + 2 \brace m + 1}$

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:
 * $\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$