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