Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m

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


 * $\displaystyle \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$

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

Proof
The proof proceeds by induction.

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

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

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

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

By definition of binomial coefficients, if $m = 0$ or $m = 1$, we have that:
 * $\dbinom 1 m = 1$

otherwise:
 * $\dbinom 1 m = 0$

Now we investigate $\displaystyle \left[{1 + 1 \atop m + 1}\right] = \left[{2 \atop m + 1}\right]$.

When $m = 0$:

When $m = 1$, from Unsigned Stirling Number of the First Kind of Number with Self we have:


 * $\displaystyle \left[{2 \atop 2}\right] = 1$

When $m > 1$, from Unsigned Stirling Number of the First Kind of Number with Greater we have:
 * $\displaystyle \left[{2 \atop 2 + k}\right] = 0$

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:
 * $\displaystyle \sum_k \left[{r \atop k}\right] \binom k m = \left[{r + 1 \atop m + 1}\right]$

from which it is to be shown that:
 * $\displaystyle \sum_k \left[{r + 1 \atop k}\right] \binom k m = \left[{r + 2 \atop m + 1}\right]$

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:
 * $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$