Original Problem:
Let $\Omega\subset \mathbb{C}$ be an open set and let $f:\Omega\to\mathbb{C}$ be holomorphic such that $f\in L^{2}(\Omega)$. Show that if $B(z,r)$, the ball of radius $r$ centered at $z$, is contained in $\Omega$, then
$$f(z) = \pi^{-1}r^{-2}\int_{B(z,r)}f(x)dx$$
Where I Am Stuck:
\begin{align} \int_{B(z,r)}f(x)dx &= \int_{B(z,r)}\sum_{n=0}^{\infty}\frac{f^{(n)}(z)(x - z)^{n}}{n!}dx\\ &= \sum_{n=0}^{\infty}\int_{B(z,r)}\frac{f^{(n)}(z)(x - z)^{n}}{n!}dx\\ &= \sum_{n=0}^{\infty}\frac{f^{(n)}(z)}{n!}\int_{B(z,r)}(x - z)^{n}dx\\ &= \int_{B(z,r)}f(z)dx + \sum_{n=1}^{\infty}\frac{f^{(n)}(z)}{n!}\int_{B(z,r)}(x - z)^{n}dx\\ &= \pi r^{2}f(z) + \sum_{n=1}^{\infty}\frac{f^{(n)}(z)}{n!}\int_{B(z,r)}(x - z)^{n}dx\\ \end{align} which is what I want as long as $\sum_{n=1}^{\infty}\frac{f^{(n)}(z)}{n!}\int_{B(z,r)}(x - z)^{n}dx = 0$.
My Question: How do I evaluate things like $\int_{B(z,r)}(x - z)^{n}dx$? I know how to integrate along a Contour in $\mathbb{C}$ but not over a disk.
Of course the concept makes sense as a Lebesgue integral but considering a limit of simple functions just seems a little too barbaric.
I feel like this is probably obvious to someone with a better understanding of complex integration than myself, and I am looking for advice on how to finish this problem.
Use polar coordinates: $x=z+\rho e^{i\theta}$, $$ \int_{B(z,r)}(x - z)^{n}dx = 2\pi \int_0^r \int_0^{2\pi}\rho^n e^{in\theta}\,\rho\,d\theta\,d\rho $$ Here $\int_0^{2\pi} e^{in\theta}\,d\theta=0$ when $n\ne 0$, because an antiderivative of $e^{in\theta}$ is $\frac{1}{in}e^{in\theta}$, which is $2\pi$-periodic.