Given that $\frac{e^z}{z^k} = z^{-k} + z^{1-k} + \frac{z^{2-k}}{2!} + \frac{z^{3-k}}{3!} + ...$ converges uniformly on any set $\{z \in C: r \leq |z| \leq Z\}$ (where $0 < r < R$), show that for $k> 0$, $\int_{|z|=1} \frac{e^z}{z^k} dz = \frac{2\pi i}{(k-1)!}$.
What I've done thus far:
$\int_{|z|=1} \frac{e^z}{z^k} dz = \int_{\gamma} (z^{-k} + z^{1-k} + \frac{z^{2-k}}{2!} + \frac{z^{3-k}}{3!} + ...)dz = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \cdot \int_{\gamma} z^{(n-1)-k}$. At this point I substitute $z = re^{i\theta}, dz = ire^{i\theta} d\theta$. We have $\sum_{k=1}^{\infty} \frac{1}{(n-1)!} \cdot \int_0^{2\pi} (re^{i\theta})^{(n-1)-k}\cdot ire^{i\theta} d\theta = \sum \frac{1}{(n-1)!} \cdot ir^{n-k} \cdot \int e^{i\theta (n-k)} d\theta$. I'm stuck with what to do here and moreover I feel like the answer is heading in the wrong direction. Tips would be appreciated.
You're on the right path, you just need to realize that $$\int e^{i\theta(n-k)}d\theta= \cases{ 0~\text{if}~n \neq k\cr2\pi~\text{if}~n = k}$$