$\int_0^{2\pi} e^{e^{i\theta}}d\theta$ This came up during my GRE prep.  Typically these questions have a trick which allows for a solution in <3 mins.  Either the ''trick'' or a hint or a worked out solution would be nice.  So far I tried expanding via EGF to little success.  I have no background in complex analysis. Thank you in advance for your time.
 A: Cauchy’s integral formula or residue theorem is unnecessary.
Recognize 
$$e^{int}=\cos (nt)+i\sin(nt)$$
$$\int_0^{2\pi}e^{int}dt=0\qquad{\forall n\in\mathbb Z\setminus\{0\}}$$
Then,
$$\int^{2\pi}_0\exp{e^{it}}dt=\int^{2\pi}_0\sum^\infty_{n=0}\frac{e^{int}}{n!}dt= \int^{2\pi}_0\frac{e^{0it}}{0!}dt+\sum^\infty_{n=1}\frac{1}{n!}\int^{2\pi}_0e^{int}dt=2\pi+0$$
The interchange of summation and integral is justified by Fubini’s theorem. 
Fubini’s theorem can be stated as

If $$\int_a^b\sum_n |f_n(x)|dx$$ converges, then $$\int_a^b\sum_n f_n(x)dx=\sum_n\int_a^b f_n(x)dx$$

NB: $|e^{int}|\equiv 1$.
A: Substitute $z = e^{i\theta}, d\theta = \frac 1{iz} \ dz$
$\oint_{|z|=1} \frac {e^z}{iz} \ dz$
Now by the Cauchy integral fromula
You will want to read this article
https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula
$\oint_\gamma \frac {f(z)}{z-a}\ dz = 2\pi i f(a)$
$2\pi i \frac {e^0}{i} = 2\pi$
A: Let $z=e^{i\theta} $ then $dz=iz\, d\theta$, hence
$$\int_0^{2\pi}e^{e^{i\theta}}\,d\theta=\int_{|z|=1}\frac{e^z}{iz}\,dz$$
Now apply Cauchy's formula. 
