$\int_0^{2 \pi} e^{e^{it} + ikt} dt = 0$ using Cauchy's integral theorem

I want to show that $$ \int_0^{2 \pi} e^{e^{it} + ikt} dt = 0 $$ for every integer $k \geq 1.$ My attempt is $$ i^k \int_0^{2 \pi} e^{e^{it} + ikt} dt = \int_0^{2 \pi} e^{e^{it}} (i e^{it})^k dt = \int_0^{2 \pi} e^{\gamma(t)} (\gamma'(t))^k dt, $$ where $\gamma : [0,2\pi] \to \mathbb{C}, \, \gamma(t) = e^{it}$. Eventually, I want to integrate over $\gamma$ and use Cauchy's integral theorem.

Any hint how to go about would be helpful, thank you.

\begin{aligned}\Large\int_{0}^{2\pi}{\operatorname{e}^{\operatorname{e}^{\operatorname{i}t}+\operatorname{i}kt}\,\mathrm{d}t}&\Large=\int_{0}^{2\pi}{\operatorname{e}^{\operatorname{e}^{\operatorname{i}t}}\left(\operatorname{e}^{\operatorname{i}t}\right)^{k-1}\operatorname{e}^{\operatorname{i}t}\,\mathrm{d}t}\\ &\Large=\frac{1}{\operatorname{i}}\oint_{\left\vert z\right\vert = 1}{z^{k-1}\operatorname{e}^{z}\,\mathrm{d}z}\end{aligned}

Since $ z\mapsto z^{k-1}\operatorname{e}^{z} $ is analytic, what can we conclude ?