On a whim, I had decided to look into ways of evaluating series of the form $$\sum_{n\ge1}\frac{1}{n^k2^n}$$ which I learned has a more general form in terms of polylogarithms: $$\text{Li}_k(x)=\sum_{n\ge1}\frac{x^n}{n^k}$$ Without prior knowledge of polylogarithms, I started working the sum where $k=1$. It's easy enough to evaluate by considering the derivative: $$F(x)=\sum_{n\ge1}\frac{x^n}{n}~~\implies~~F'(x)=\frac{1}{x}\sum_{n\ge1}x^n$$ and so on, which tells me the sum is $\ln2$.
I noticed that I could keep taking the derivative in this fashion for larger values of $k$ and came up with the recursive derivative relation for the polylog (equation (18) in the MathWorld link). If I denote $f_k\equiv\text{Li}_k(x)$, then $$\frac{df_k}{dx}=\frac{f_{k-1}}{x}$$ I considered successive orders of the derivative: $$\frac{d^2f_k}{dx^2}=\frac{-f_{k-1}+f_{k-2}}{x^2}\\ \frac{d^3f_k}{dx^3}=\frac{2f_{k-1}-3f_{k-2}+f_{k-3}}{x^3}\\ \frac{d^4f_k}{dx^4}=\frac{-6f_{k-1}+11f_{k-2}-6f_{k-3}+f_{k-4}}{x^4}$$ I hadn't known about the Stirling numbers before this. As it turns out, the coefficients of the numerator's terms belong to this sequence.
More generally, it would appear that $$\frac{d^mf_k}{dx^m}=\frac{1}{x^m}\sum_{i=1}^ms(m,i)f_{k-i}$$ Is this result at all surprising?
Edit: For another thing, is it at all surprising that $$\sum_{i=1}^m s(m,i)=0~~?$$