Over here we had a similar looking question:
and the answer was that (rather unremarkably) we could remove the singularity made by $1/(e^z-1)$ for $z=0$ and conclude Res$((z^n)/(e^z-1))=0$ for all $n>0$ and Res$(1/(e^z-1))=1$.
I think this residue calculation becomes rather more interesting when we put a $-$ in front of the $n$. Then we have an (n+1)th order pole at $z=0$. Can we calculate the residue other than with this limit formula?