Proof
(related to Proposition: Inversion Formulas For Stirling Numbers)
- 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 $$x^{n}= \sum_{r=1}^n\left\{\begin{array}{c}n\\r\end{array}\right\}x^\underline{r}.$$
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References
Bibliography
- Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition
