I want to find the residue of $f(z)=z^me^{1/z}$ at $0$. Could you please check if this is correct?
$z^me^{1/z }=z^m\sum_{k=0}^{\infty}\frac{(\frac{1}{z})^k}{k!}=z^m\sum_{k=0}^{\infty}\frac{1}{k!}z^{-k}=\sum_{k=0}^{\infty}\frac{1}{k!}z^{m-k}$
And at $k=m+1$ we get $z^{-1}$ so the residue of the function is $\frac{1}{(m+1)!}$ since it's the coefficient of $(z-0)^k$?
