I was trying to integrate the contour integral $$\int_\gamma \frac{\vert z \vert e^z}{z^2}$$ where $\gamma$ parametrizes the unit circle counterclockwise. I cannot use the Generalized Cauchy Integral Formula since the numerator of the integrand is not holomorphic, and I cannot use the residue calculus since the integrand is not holomorphic. So my next idea was to use the fact that since $\gamma(t)=e^{2\pi it}$, where $t\in \lbrack0,1\rbrack$ we have $$\int_\gamma\frac{\vert z \vert e^z}{z^2}=\int_0^1\frac{e^{e^{2\pi it}}}{e^{4\pi it}}2\pi ie^{2\pi it}dt=2\pi i\int_0^1\frac{e^{e^{2\pi it}}}{e^{2\pi it}}dt$$
WolframAlpha claims that $$\int_0^1\frac{e^{e^{2\pi it}}}{e^{2\pi it}}dt=1$$ but I'm not sure how to verify this. I tried using a U-substitution $u=e^{2\pi it}$, but then the integral becomes $0$ as the limits of the integration both become $1$ after the substitution.
How do I show $$\int_0^1\frac{e^{e^{2\pi it}}}{e^{2\pi it}}dt=1$$ and does the U-substitution method work for complex integrals?
Hint: This is easier than it looks. On the unit circle, $|z|=1$.