Recently I read in lecture notes that for $\alpha \in \mathbb{N}^m$ with $\vert\alpha\vert = r$ the following identity holds: $$ \sum_{} \frac{1}{\alpha!} = \frac{m^r}{r!}.$$
Appearently one can prove that by induction. I tried but did not succeed. How to prove this correctly?
On
Seems like an induction on $m$. Suppose the result holds for $\Bbb N^m$. For $\alpha\in\Bbb N^{m+1}$write $$\alpha=(\alpha',\beta),$$where $\alpha'\in\Bbb N^m$ and $\beta\in\Bbb N$. Then $$\sum_{\alpha\in\Bbb N^{m+1},|\alpha|=r}\frac{1}{\alpha!} =\sum_{\beta=0}^r\frac1{\beta!}\sum_{\alpha'\in\Bbb N^m,|\alpha'|=r-\beta} \frac{1}{(\alpha')!}=\sum_{\beta=0}^r\frac{m^{r-\beta}}{\beta!(r-\beta)!} =\frac{(m+1)^r}{r!}.$$
Your identity is a simple consequence of the multinomial theorem. For $x=(x_1,\ldots,x_n)$ and $\alpha=(\alpha_1,\ldots,\alpha_m)\in\mathbb{N}^m$ we write $x^\alpha:=x_1^{\alpha_1}\cdots x_m^{\alpha_m}$ (and, as you do, $|\alpha|:=\alpha_1+\cdots+\alpha_m$, $\alpha!:=\alpha_1!\cdots\alpha_m!$). Then \begin{equation*} (x_1+\cdots+x_m)^r \,=\, \sum_{\alpha\in\mathbb{N}^m\!,\,|\alpha|=r}\frac{r!}{\alpha!}\,x^\alpha~. \end{equation*} You obtain your identity taking $x_1=\cdots=x_m=1$. Now of course you must prove the multinomial theorem. You will find that proving the multinomial theorem requires precisely as much effort and ingenuity as proving your very special identity. Or, you may simply accept the multinomial theorem as well-known.