Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial

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


 * $\displaystyle \sum_{k \mathop \le n} \left[{k \atop m}\right] \frac {n!} {k!} = \left[{n + 1 \atop m + 1}\right]$

where:
 * $\displaystyle \left[{k \atop m}\right]$ denotes an unsigned Stirling number of the first kind
 * $ n!$ denotes a factorial.

Proof
The proof proceeds by induction on $m$.

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

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:
 * $\displaystyle \sum_{k \mathop \le n} \left[{k \atop r}\right] \frac {n!} {k!} = \left[{n + 1 \atop r + 1}\right]$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop \le n} \left[{k \atop r + 1}\right] \frac {n!} {k!} = \left[{n + 1 \atop r + 2}\right]$

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:
 * $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} \left[{k \atop m}\right] \frac {n!} {k!} = \left[{n + 1 \atop m + 1}\right]$