Proving that $\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{it})|^2dt=\sum_{n=0}^{\infty}|a_n|^2r^{2n}$

$$ \int_0^{2\pi}| f(re^{it})|^2 dt = \int_0^{2\pi}f(re^{it})\overline{f(re^{it})}dt = \int_0^{2\pi}\sum_{n=0}^{\infty}a_nr^ne^{int}\sum_{m=0}^{\infty}\overline{a_m}r^me^{-imt} dt $$

Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have

$$ \frac{1}{2\pi}\int_0^{2\pi}| f(re^{it})|^2 dt = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_n\overline{a_m}r^{n+m}\int_0^{2\pi}e^{i(n-m)t}dt = \frac{1}{2\pi}\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_n\overline{a_m}r^{n+m} 2\pi\delta_{nm} = \sum_{n=0}^{\infty}|a_n|^2r^{2n}.$$

As for how to so that $\int_0^{2\pi}e^{i(n-m)t}dt = 2\pi\delta_{nm}$, the proof is as follows:

Suppose $n = m$ then we wish to evaluate $\int_0^{2\pi}e^{i(n-m)t}dt = \int_0^{2\pi}e^{0}dt = \int_0^{2\pi} 1 dt = 2\pi$.

Suppose $n\neq m$ then we have $\int_0^{2\pi}e^{i(n-m)t} dt = \frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2\pi} = \frac{1}{i(n-m)}\left(e^{2\pi i(n-m)} - 1\right)$. But since $2\pi i(n-m)$ is a multiple of $2\pi$, $e^{2\pi i(n-m)} = 1$ and so it evaluates to $0$. QED.