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:
 * $\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$

where:
 * $\displaystyle {n \brack k}$ 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 $\map P n$ be the proposition:
 * $\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

Basis for the Induction
$\map P 2$ is the case:

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\displaystyle \sum_k \paren {-1}^k {m \brack k} = \delta_{m 0} - \delta_{m 1} = 0$

from which it is to be shown that:
 * $\displaystyle \sum_k \paren {-1}^k {m + 1 \brack k} = \delta_{\paren {m + 1} 0} - \delta_{\paren {m + 1} 1} = 0$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$