Alternative method to find this infinite sum? 
Find where m is an integer:
  $$S = 1+\sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$
When we rewrite this sum as one plus a function of $x$ and multiply
  the $n^{th}$ term of the series by $x^{mn-1}$, the resulting function
  solves for the differential equation $$\frac{d^{m-1}f(x)}{dx^{m-1}} = \frac1m + \frac1{xf(x)}$$ and hence we can replace $f(x)$ with $s(x)$ and set
  $s(0) = 1$. Then use that to find $s(1)$ which is $S$.

I'm looking for an alternative method because I'm having trouble dealing with this differential equation.

Considering $$S_m = \sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$ beside the case $m=2$ for which $$S_2=\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \text{erf}\left(\frac{1}{2}\right)$$ as Patrick Stevens commented, we are left with hypergeometric functions which apparently do not simplify at all.
They show quite nice patterns 
$$S_3=\frac{1}{3!} \, _1F_2\left(1;\frac{4}{3},\frac{5}{3};\frac{1}{3 ^3}\right)$$
$$S_4=\frac{1}{4!} \,
   _1F_3\left(1;\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{1}{4^4}\right)$$
$$S_5=\frac{1}{5!} \,
   _1F_4\left(1;\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{1}{5^5}\right)$$
$$S_6=\frac{1}{6!} \,
   _1F_5\left(1;\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac{
   1}{6^6}\right)$$
$$S_7=\frac{1}{7!}\,
   _1F_6\left(1;\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7},\frac{13}{7};\frac{1}{7^7}\right)$$
In fact, it seems that the asymptotics is just $S_m\sim\frac{1}{m!}$.
