You do not even need Cauchy closed curve theorem here. Just note that $\displaystyle \left( e^z \right)' = e^z$ which means if $\displaystyle \Gamma$ it a piecewise smooth closed curve then $\displaystyle \oint_{\Gamma} e^z dz = 0$.

You prove this by writing,

$\displaystyle 0=\oint \limits_{|z| = 1} e^z dz = \int_0^{2\pi} e^{e^{i\theta}} ie^{i\theta} d\theta \implies \int_0^{2\pi} e^{e^{i\theta}} e^{i\theta} = 0$

Now split the real and imaginary parts.

Hint: $\displaystyle e^{e^{i\theta}} = e^{\cos \theta} \cos (\sin \theta) + i e^{\cos \theta}\sin (\sin \theta)$.