Inspired by Dobinski formula, by lucky guess I find, that (for natural $a,b$) $$f(a,b)=e^{-1/a}\sum\limits_{n=0}^{\infty}\frac{(an)^{b}}{a^{n}n!}=\sum\limits_{k=1}^{b}{b\brace k}a^{b-k}$$ but I have problems with $$f(a,b,c)=e^{-c^a/a}\sum\limits_{n=0}^{\infty}\left(\frac{c^a}{a}\right)^{n}\frac{(an)^{b}}{n!}$$ For example, for $a=1$, $b=1,2,\cdots$, $c=\frac{3}{2}$, we have $$\frac{3\cdot1}{2^1}, \frac{3\cdot5}{2^2}, \frac{3\cdot31}{2^3}, \frac{3\cdot227}{2^3}, \frac{3\cdot1897}{2^4}, \frac{3\cdot17693}{2^5}, \cdots$$ Why only denominator's degree growth?
How can I find it in general?
We have that $$ \eqalign{ & \sum\limits_{0\, \le \,n} {x^{\,n} {{n^{\,b} } \over {n!}}} \quad \left| {\,0 \le b \in Z} \right.\quad = \sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}\sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{ b \hfill \cr k \hfill \cr} \right\}n^{\,\underline {\,k\,} } } } = \cr & = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{ b \hfill \cr k \hfill \cr} \right\}\sum\limits_{0\, \le \,n} {n^{\,\underline {\,k\,} } {{x^{\,n} } \over {n!}}} } = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{ b \hfill \cr k \hfill \cr} \right\}x^{\,k} \sum\limits_{0\, \le \,n} {n^{\,\underline {\,k\,} } {{x^{\,n - k} } \over {n!}}} } = \cr & = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{ b \hfill \cr k \hfill \cr} \right\}x^{\,k} {{d^{\,k} } \over {dx^{\,k} }}\sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}} } = e^{\,x} \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{ b \hfill \cr k \hfill \cr} \right\}x^{\,k} } \cr} $$ then you can easily adapt it to your case.