First Inversion Formula for Stirling Numbers

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


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

where:
 * $\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind
 * $\displaystyle \left\{ {k \atop m}\right\}$ 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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$

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

Thus $P \left({0}\right)$ has been shown to hold.

This is the basis for the induction.

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

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

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

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

Also see

 * Second Inversion Formula for Stirling Numbers