what it says on the tin tbh
if we have an $f$ holomorphic on $\Omega$, then for all $z_0 \in \Omega$ we can find some power series expansion centered on it with a nonzero radius of convergence:
\[f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\](above formula only true in some ball around $z_0$ i spose)
this proof does need the cauchy integral theorem though, just so you know
it starts like most, by fixing a $z_0 \in \Omega$ and a $0 < R$ s.t. $B_R(z) \subseteq \Omega$
where the cauchy integral theorem comes in is that $\forall z \in B_R(z_0)$ we can write
\[f(z)=\frac{1}{2\pi i}\int_{|\zeta-z_0|=R} \frac{f(\zeta)}{\zeta-z}d\zeta\]the kicker is that since $z$ is closer to $z_0$ than $\zeta$ for all $\zeta$, we can do some geometric series crap to write
\[\begin{align*} \frac{1}{\zeta-z} &= \frac{1}{(\zeta-z_0)-(z-z_0)} \\ &= \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}} \end{align*}\]you can verify the second equality by using the geo series formula, trust
then we plug this back in to that integral expression for $f(z)$ earlier:
\[\begin{align*} f(z) &= \frac{1}{2\pi i}\int_{|\zeta-z_0|=R} \frac{f(\zeta)}{\zeta-z}d\zeta \\ &= \frac{1}{2\pi i}\int_{|\zeta-z_0|=R} f(\zeta) \cdot \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}} d\zeta \\ &= \frac{1}{2\pi i}\sum_{n=0}^\infty \int_{|\zeta-z_0|=R} f(\zeta) \cdot \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}} d\zeta \\ &= \frac{1}{2\pi i}\sum_{n=0}^\infty (z-z_0)^n \int_{|\zeta-z_0|=R} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} d\zeta \end{align*}\]the fact that you can swap the integral and the infinite sum is actually nontrivial here. my prof explained it with some stuff from real analysis, and i'm not sure if there's any other way to justify it
the idea is that the terms as a function of $\zeta$ (the variable of integration) converge uniformly in the summation, so the integral of the result is equivalent to the sum of all the tiny little integrals as well
now yeah the order of the terms is kinda wonky but this is certainly a power series expansion around $z_0$! $\square$