Computing a complex integral with residue theorem

Question

I'm trying to compute: $$ \int_C z^n e^{1/z}dz, \quad \quad C=\{z \in \mathbb{C}: |z|=1\}. $$ I wanna use the residue theorem. For that, I did $$ f(z)=z^ne^{1/z}=z^n\sum_{k=0}^{\infty}\frac{z^k}{k!}=\sum_{k=0}^{\infty}\frac{z^{n-k}}{k!}=z^n+z^{n-1} +\frac{z^{n-2}}{2!}+\cdots $$ So, for $n \leq -2$, the of $f(z)$ is zero since there is not the $a_{-1}$ term. For $n \in\{-1,0,1,2...\}$, the residue is $1/k!$. Hence, by residue theorem, we get $$ \int_C z^n e^{1/z}dz=\frac{2 \pi i}{(n+1)!} $$ Just wanna make sure I'm doing well my reasoning. Thanks.