I came across the series $$A(N):=\sum_{k=1}^\infty \frac{k^N}{2^k},~~N\in\mathbb{N},N\geq 4,$$ which is the polylogarithm of order $-N$ evaluated at $\frac{1}{2}$.
I got the hint that this can be bounded in terms of the $N-$th moment of an exponentially distributed random variable $X$ with parameter $\ln(2)$.
Indeed, if I approximate the corresponding integral by a Riemann- sum, something close to the target expression turns up: For $B>0$, we have
\begin{eqnarray*} \mathbb{E}[X^N]&=&\ln(2)\int_{0}^\infty x^N2^{-x}\ \mathrm{d}x\\ &\geq&\ln(2)\int_{0}^B x^N 2^{-x}\ \mathrm{d}x\\ &=&\ln(2)\lim_{n\rightarrow\infty,n\geq B}\frac{B}{n}\sum_{k=1}^n\frac{\left(k\frac{B}{n}\right)^N}{2^{k\frac{B}{n}}}\\ &\geq&\ln(2)\lim_{n\rightarrow\infty,n\geq B}\left(\frac{B}{n}\right)^{N+1}\sum_{k=1}^n\frac{k^N}{2^{k}}. \end{eqnarray*} So far I couldn't get closer to the solution. Do you have an idea?
Of course, I am also open to any other approach for finding the desired bound, i.e. a bound of the form $$A(N)\leq c^N N!$$ For example, I tried the representation of the polylogarithm that uses the Stirling numbers of the second kind, but didn't get as ``far'' as above.