First Inversion Formula for Stirling Numbers

Theorem
For all $m, n \in \Z_{\ge 0}$:


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

where:
 * $\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
 * $\ds {k \brace m}$ denotes a Stirling number of the second kind
 * $\delta_{m n}$ denotes the Kronecker delta.

Proof
The proof proceeds by strong induction.

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

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ has been shown to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P j$ is true for all $j$ such $0 \le j \le r$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \forall j: 0 \le j \le r: \forall m \in \Z_{\ge 0}: \sum_k {j \brack k} {k \brace m} \paren {-1}^{j - k} = \delta_{m j}$

from which it is to be shown that:
 * $\ds \forall m \in \Z_{\ge 0}: \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k} = \delta_{m \paren {r + 1} }$

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

Also see

 * Second Inversion Formula for Stirling Numbers