Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:
 * $\displaystyle \sum_k \left({-1}\right)^k \left[{n \atop k}\right] = \delta_{n 0} - \delta_{n 1}$

where:
 * $\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind
 * $\delta_{n 0}$ denotes the Kronecker delta.

Proof
The proof proceeds by induction on $n$.

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

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

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

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

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction
$P \left({2}\right)$ is the case:

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_k \left({-1}\right)^k \left[{m \atop k}\right] = \delta_{m 0} - \delta_{m 1} = 0$

from which it is to be shown that:
 * $\displaystyle \sum_k \left({-1}\right)^k \left[{m + 1 \atop k}\right] = \delta_{\left({m + 1}\right) 0} - \delta_{\left({m + 1}\right) 1} = 0$

Induction Step
This is the induction step:

So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \left({-1}\right)^k \left[{n \atop k}\right] = \delta_{n 0} - \delta_{n 1}$