Consider a region $\omega \subset \mathbb{C}$, and a holomorphic function $f : \omega \rightarrow \mathbb{C}$. I wish to prove that if $D$ is an open disk with center $z_{0}$ whose closure is contained in $w$, then there exists $(a_{n})_{n \in \mathbb{N}}$ such that
$f(z) = \displaystyle\sum_{n = 0}^{\infty} a_{n}(z - z_{0})^{n}$, for $z \in D$.
To show this, we set $a_{n} = \displaystyle\frac{f^{(n)}(z_{0})}{n!}$. Then, plugging in to the above sum gives
$\displaystyle\sum_{n = 0}^{\infty} \frac{(z - z_{0})^{n}}{2\pi i} \int_{C} \frac{f(w)}{(w - z_{0})^{n + 1}}\ dw$, after applying Cauchy's formula for $f^{(n)}(z_{0})$ (where $C$ is the boundary of $D$).
If I could justify changing the order of the summation and integral signs, it would be pretty easy to continue and show that this equals $f(z)$. The book (Stein's complex analysis) follows a different argument to prove this, but it also switches order once, saying only that it's valid because of uniform convergence. So does this converge uniformly?