Theorem
Let $\Omega\subseteq \mathbb{C}$ be open and $f\in H(\Omega)$ ($f$ analytic on $\Omega$). If $ C(z_{0},R)\subseteq\Omega$ (where $C(z_0,R)$ is the circle with origin $z_0$ and radius $R$), then we can represent $f$ on $C(z_{0},R)$ as a power series with convergence radius $\geq R$.
Proof
Let $0<r<R$ and $\gamma:[0,2\pi]\rightarrow\Omega$ be the path $\gamma(t)=z_{0}+re^{it}$.
Because $C(z_{0},R)$ is convex and $\gamma$ is in this circle, we have the following from the Cauchy integral formula:
$$
f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in C(z_{0},\ r))
$$
$\color{red}{\text{(1) Why not directly on } C(z_{0},\ R)? }$
Because $C(z_{0},\ r)\subseteq \mathbb{C}\backslash \gamma^{*}$ it follows that $f(z)=\displaystyle \sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ for a power series with convergence radius $\geq r$. Because $ r\in (0,\ R)$ is chosen arbitrarily and because the the coefficients $c_{n}$ are determined by the differentials $f^{(n)}(z_{0})$, the power series $\displaystyle \Sigma_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ has the convergence radius $\geq R$.
$\color{red}{\text{(2) Why are the coefficients calculated via }f^{(n)}(z_{0})? }$
$\color{red}{\text{(3) Why does it follow that the convergence radius is } \geq R? }$
Best Answer
For (1), we choose $r<R$ because the path $\gamma_R(t)=z_0+Re^{it}$ for $t\in[0,2\pi]$ may not lie inside the region $\Omega$. This is because while $C(z_o,R)\subset\Omega$, its boundary, which is $\gamma_R^*$, may not be contained within $\Omega$. Now, to see why the integral formula only holds when $z\in C(z_0,r)$, and not on its boundary (which is $\gamma^*$ in your statement), this is simply due to the statement of the Cauchy Integral Formula. That is, it only holds for points that do not lie on $\gamma$.
For (2), the coefficients $c_n$ are determined by the differentials $f^{(n)}(z_0)$ because of the Taylor formula (which we can obtain from the finite Taylor formula if need be).
For (3), we have shown that the power series has radius of convergence $r$ where $r$ is an arbitrary positive number less than $R$. Thus, the radius of convergence is at least $\sup_{0<r<R} r= R$. So the radius of convergence is greater than or equal to $R$.