Compute \begin{equation}I=\frac{1}{2\pi i}\int_{C(0,1)}z^n\exp\left(\frac{2}{z}\right)\textrm{d}z\end{equation} where $C(0,1)$ is the unit circle centred at $0$ oriented anticlockwise, for integer values $n$.
My solution: Note the Laurent expansion of $\exp(z)=1+z+\frac{z^2}{2!}+\cdots$, hence $\exp\left(\frac{2}{z}\right)=1+\frac{2}{z}+\cdots +\frac{z^k}{k!2^k}+\cdots $. Hence we have that $z^n\exp\left(\frac{2}{z}\right)=z^n+2z^{n-1}+\cdots +\frac{z^{k+n}}{k!2^k}$. Then to determine the residue we need only look at the coefficient of $1/z$, for a particular $n$. Is this correct?
Let $$f(z)=z^ne^{\frac {2}{z}}$$
$$z^ne^{\frac {2}{z}}= \sum_{m=0}^\infty 2^m\frac{z^{(-m+n)}}{m!}(*)$$
By definition $Res(f,0)=a_{-1}$ of the Laurent series $(*)$. So we want $-m+n=-1\implies m=n+1\implies \operatorname{Res}(f,0)=a_{-1}=\frac{2^{n+1}}{(n+1)!}$