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

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


 * $\ds \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$

where:
 * $\dbinom n m$ denotes a binomial coefficient
 * $\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
 * $\ds {n + 1 \brace k + 1}$ denotes a Stirling number of the second kind.

Proof
The proof proceeds by induction on $n$.

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

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:
 * $\ds \sum_k {r + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom r m$

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

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:
 * $\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$

Also see

 * Sum over $k$ of $\ds {n + 1 \brack k + 1}$ by $\ds {k \brace m}$ by $\paren {-1}^{k - m}$