Let $n,m\ge 0$ be integers. The Stirling numbers of the first and second kind obey the following inversion formulas:
$$\begin{align} \sum_{k=0}^\infty \left[\begin{array}{c}n\\k\end{array}\right]\left\{\begin{array}{c}k\\m\end{array}\right\}(-1)^{n-k}&=[m=n]\nonumber,\\ \sum_{k=0}^\infty \left\{\begin{array}{c}n\\k\end{array}\right\}\left[\begin{array}{c}k\\m\end{array}\right](-1)^{n-k}&=[m=n]\nonumber. \end{align}$$ Above, the Iverson notation $[m=n]$ is used and means that the sums equal $1$ if and only if $m=n$, otherwise they equal $0.$ Please note that the above sums are, in fact, finite, since all summands are $0$ if the index $k$ is above $\min(m,n)$.
Proofs: 1
