Let $\Omega\subseteq\mathbb{C}$ be a connected open set, then $f:\Omega\rightarrow \mathbb{C} $ is analytic if for every $z_0\in\Omega$ there are $r>0$ and $(a_n)\subseteq\mathbb{C}$ such that $f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$ for every $z\in B(z_0,r)\subseteq\Omega $, and in fact $ a_n=\frac{f^{(n)}(z_0)}{n!}$.
How can we show that given some $ z_0\in\Omega$ the equality $f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ actually holds for every $z\in\Omega$ and not just in some neighborhood of $z_0$?
You can not show this equality ! Example $ \Omega=\mathbb C \setminus \{1\}$, $f(z)=\frac{1}{1-z}$ and $z_0=0$.
We have $f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n=\sum_{n=0}^\infty z^n$ only for $z \in \Omega$ with $|z|<1$.
Remark: if $ \Omega=\mathbb C$ and if $f$ is entire, then we have for each $z_0 \in \mathbb C$ that $f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ for all $z \in \mathbb C$.