Let $\gamma\colon [0,2\pi]\longrightarrow \mathbb{C},\ t\mapsto e^{it}$ be a path. I wanted to compute $\int_\gamma e^{1/z}\mathrm{d}z$ by using the series form of the exponential function.
So put $\int_\gamma e^{1/z}\mathrm{d}z=\int_\gamma \lim_{n\to\infty} \sum_{j=0}^n 1/(j!z^j)\mathrm{d}z$. Because, the series converges uniformly and the integral is additive, I can write $\lim_{n\to\infty}\sum_{j=0}^n\int_\gamma 1/(j!z^j)\mathrm{d}z$
Then substituting the path yields: $\lim_{n\to\infty}\sum_{j=0}^n\int_0^{2\pi} i/j! e^{-it(j-1)}\mathrm{d}t$ and all terms are zero except at $j=1$ which gives $2\pi i $ for the whole integral.
I have not learned the residual theorem or cauchy integral yet, so I can not give a more compact answer on how to integrate. Can somebody confirm my answer (hopefully much faster) by using more advanced theorems?