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

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


 * $\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop m}\right\} \left({m + 1}\right)^{n - k} = \left\{ {n + 1 \atop m + 1}\right\}$

where $\displaystyle \left\{ {k \atop m}\right\}$ etc. denotes a Stirling number of the second kind.

Proof
The proof proceeds by induction on $m$.

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

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:
 * $\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop r}\right\} \left({r + 1}\right)^{n - k} = \left\{ {n + 1 \atop r + 1}\right\}$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop r + 1}\right\} \left({r + 2}\right)^{n - k} = \left\{ {n + 1 \atop r + 2}\right\}$

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:
 * $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} \left\{ {k \atop m}\right\} \left({m + 1}\right)^{n - k} = \left\{ {n + 1 \atop m + 1}\right\}$