Is there a nice way to write the sum $\sum_{\sum k_i = m} \frac{1}{k_1! k_2! ... k_n!}$? Is there any nice way to write $$\sum_{\sum k_i = m} \frac{1}{k_1! k_2!  ...  k_n!}$$
where $m$ and the $k_i$ are positive integers and $n$ is a fixed positive integer? I thought maybe multiplying by $m!$ and trying to get to a binomial sum, but I'm not sure.
This comes up in dealing with independent trials from a poisson distribution, namely when working with the statistic which sums the values of the trials. 
 A: Note that our primary tool here will be the series expansion: 
$$e^x=\sum_{k\ge 0}\frac{1}{k!}x^k\tag{1}$$
Then let us consider the sum 
$$\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}$$
for two distinct cases: 


*

*For all $k_i\ge 0$.

*For all $k_i\ge 1$.



Case 1: $k_i\ge 0$
We see, by raising $(1)$ to the power $n$, that
$$(e^{x})^n=\sum_{m\ge 0}\left(\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}\right)x^m\, ,$$
and thus taking the $x^m$ coefficient:

$$[x^m]e^{nx}=\frac{n^m}{m!}=\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}\tag{Answer 1}$$


Case 2: $k_i\ge 1$
We want to do something similar to case 1 but avoiding $k$ values of $0$, hence consider the modification of $(1)$:
$$e^x-1=\sum_{k\ge 1}\frac{1}{k!}x^k\tag{2}$$
and raising $(2)$ to power $n$
$$(e^x-1)^n=\sum_{m\ge n}\left(\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}\right)x^m$$
then taking the coefficient of $x^m$
$$[x^m](e^x-1)^n=\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}\, .$$
We can write $(e^x-1)^n$ using the binomial expansion
$$(e^x-1)^n=\sum_{j=0}^{n} \binom{n}{j}(-1)^{n-j}e^{jx},$$
and therefore
$$[x^m](e^x-1)^n=\sum_{j=0}^{n}\binom{n}{j}(-1)^{n-j}[x^m]e^{jx}$$
$$\implies [x^m](e^x-1)^n=\sum_{j=0}^{n}\binom{n}{j}(-1)^{n-j}\frac{j^m}{m!}\, .$$
So, we have:
$$\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}=\frac{1}{m!}\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^m\, .$$
This is related to the Stirling numbers of the second kind:
$${m\brace n}=\frac{1}{n!}\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^m$$
by

$$\sum_{\sum k_i=m}\frac{1}{k_1!\cdots k_n!}=\frac{n!}{m!}{m\brace n}\tag{Answer 2}$$

A: The Multinomial Theorem says,
$$
\sum_{\sum\limits_{i=1}^nk_i=m}\frac{m!}{k_1!\,k_2!\,\dots\,k_n!}1^{k_1+k_2+\cdots+k_n}=(\overbrace{1+1+\cdots+1}^{n\text{ copies}})^m\tag1
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{\sum\limits_{i=1}^nk_i=m}^{\vphantom{1}}\frac1{k_1!\,k_2!\,\dots\,k_n!}=\frac{n^m}{m!}}\tag2
$$
Note that if we add $(2)$ for all $m\ge0$, we get $e^n$ which makes sense because the sum then allows all $k_1,k_2,\dots,k_n$ giving
$$
\left(\sum_{k=0}^\infty\frac1{k!}\right)^n\tag3
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k_{1} = 0}^{\infty}\ldots\sum_{k_{1} = 0}^{\infty}
{1 \over k_{1}!\cdots k_{n}!}\,\bracks{z^{m}}z^{k_{1} + \cdots + k_{n}} =
\bracks{z^{m}}\pars{\sum_{k = 0}^{\infty}{z^{k} \over k!}}^{n} =
\bracks{z^{m}}\pars{\expo{z}}^{n} =
\bracks{z^{m}}\expo{nz} = \bbx{n^{m} \over m!}
\end{align}
A: Combinatorial interpretation:
the fraction $\frac{m!}{k_1!\cdots k_n!}$ is the number of ways to partition $m$ distinct balls to $n$ distinct cells so that the $i$-th cell gets $k_i$ balls.
Summing this over all $k_1,\ldots,k_n$ we get the number of all surjections $\{1,\ldots,m\}\rightarrow\{1,\ldots,n\}$, i.e. $n!{m\brace n}$. Dividing back by $m!$, we get the answer.
