# Proof

• By hypothesis, $n,m\ge 0$ are integers.
• Using the lemma for the stirling numbers and rising factorial powers and because $\left\{\begin{array}{c}k\\m\end{array}\right\}=0$ for $m > k,$ we get \begin{align} x^\overline{n}&= \sum_{k=1}^n\left[\begin{array}{c}n\\k\end{array}\right]x^k\nonumber\\ &= \sum_{k=1}^n\left[\begin{array}{c}n\\k\end{array}\right]\sum_{r=1}^k\left\{\begin{array}{c}k\\r\end{array}\right\}(-1)^{k-r}x^\overline{r}\nonumber\\ &= \sum_{k=1}^n\left[\begin{array}{c}n\\k\end{array}\right]\sum_{m=1}^n\left\{\begin{array}{c}k\\m\end{array}\right\}(-1)^{k-m}x^\overline{m}\nonumber\\ &= \sum_{k=1,m=1}^n\left[\begin{array}{c}n\\k\end{array}\right]\left\{\begin{array}{c}k\\m\end{array}\right\}(-1)^{n-k}x^\overline{m}\nonumber\\ \end{align}
• Because this holds for all $x$, it follows that the coefficients of the sum on the right are all $0$ except those, for which $m=n.$
• Thus, using the Iverson notation and since all summands for indices $k$ exceeding $\min(m,n)$ equal $0$, we can write $$\sum_{k=0}^\infty \left[\begin{array}{c}n\\k\end{array}\right]\left\{\begin{array}{c}k\\m\end{array}\right\}(-1)^{n-k}=[m=n].$$
• The proof of the other identity $$\sum_{k=0}^\infty \left\{\begin{array}{c}n\\k\end{array}\right\}\left[\begin{array}{c}k\\m\end{array}\right](-1)^{n-k}=[m=n]$$ can be done analogously plugging the defining equation of Stirling numbers of the first kind $$x^\underline{n}= \sum_{r=1}^n\left[\begin{array}{c}n\\r\end{array}\right](-1)^{n-r}x^r$$ into the equation of the Stirling numbers of the second kind $$x^{n}= \sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\}x^\underline{r}.$$

Github: ### References

#### Bibliography

1. Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition