Why is $\frac{1}{n!}=\frac{1}{2\pi r^n}\int_{0}^{2\pi}e^{re^{it}}e^{-int}dt$

Question

How to prove that $\displaystyle\frac{1}{n!}=\frac{1}{2\pi r^n}\int_{0}^{2\pi}e^{re^{it}}e^{-int}dt$ for any natural number $n$ and poisitive real number $r$

I got with $f(z)=e^z$ and Cauchy's Integral formula;

$\displaystyle\int_{0}^{2\pi}e^{re^{it}}e^{-int}dt=\int_{|z=1|}\frac{e^z}{z}dz=\int_{|z|=1}\frac{f(z)}{(z-0)^{((n-1)+1)}}dz=\frac{2\pi i}{(n-1)!}f^{(n-1)}(0)=\frac{2\pi i}{(n-1)!}$

so something went wrong, can you help ?

Answer 1

For any $m\in\mathbb{N}$ we have: $$\int_{0}^{2\pi}e^{imt}e^{-int}\,dt = 2\pi\, \delta_{m,n}$$ and since the exponential function is an entire function satisfying: $$ e^z = \sum_{k\geq 0}\frac{z^k}{k!}$$ we have: $$ \int_{0}^{2\pi}e^{r e^{it}}e^{-nit}\,dt = \int_{0}^{2\pi}\frac{r^n e^{nit}}{n!}e^{-nit}\,dt = \frac{2\pi\, r^n}{n!}$$ as wanted.