I'm looking for ways to compute the coefficients of the power series $$ \sum_{n=0}^\infty\frac{(-1)^n}{n!^s}=\sum_{k=0}^{\infty}c_k s^k $$ (a prior version of the question asked whether such an expansion exists at all, at least in the asymptotic sense as $s\to0^+$; thanks to the answer by @Svyatoslav, now I know that it does).
For now, I know that $\color{blue}{c_0=\frac12}$ and $\color{blue}{c_1=\frac14\log\frac2\pi}$. The value of $c_0$ is obtained e.g. in this answer, and I've just used this approach to get the value of $c_1$. Roughly, for $f(s)=\sum_{n=0}^\infty(-1)^n e^{-\lambda_n s}$, we define $$ f_0(t)=\sum_{n=0}^\infty\mathbf{1}_{(\lambda_{2n},\lambda_{2n+1})}(t),\qquad f_{n+1}(t)=\int_0^t f_n(x)\,dx $$ and, under certain conditions, if we have $f(s)\asymp\sum_{k=0}^{(\infty)}c_k s^k$ as $s\to 0^+$, then $$ c_n=\lim_{t\to\infty}\left(\frac{f_{n+1}(t)}{t}-\sum_{k=0}^{n-1}\frac{c_k t^{n-k}}{(n-k+1)!}\right). $$ We replace $t$ by $\lambda_m$ (and $m\to\infty$) and use $f_n(\lambda_m)=(1/n!)\sum_{k=0}^{m-1}(-1)^k(\lambda_m-\lambda_k)^n$.
With $\lambda_n=\log n!$, (even) the computation of $c_1$ this way results in a limit with sums of logarithmic terms, which is not too hard, but a little bit tedious, to evaluate. And $c_2$ is out of my patience for now ;)