Let \(f\) be holomorphic on an open set containing the closed disc \(\overline{D_R(p)}\), and write \(f(z)=\sum_{k\ge0}c_k(z-p)^k\) near \(p\). Then for every \(k\ge0\), \[ c_k=\frac{f^{(k)}(p)}{k!}=\frac{1}{2\pi i}\int_{\abs{z-p}=R}\frac{f(z)}{(z-p)^{k+1}}\dz=\frac{1}{2\pi R^k}\int_0^{2\pi}f(p+Re^{i\theta})e^{-ik\theta}\dtheta. \] Consequently, with \(M_R\coloneqq\max_{\abs{z-p}=R}\abs{f(z)}\), \[ \abs{c_k}\le\frac{M_R}{R^k}, \] so \(\limsup_k\abs{c_k}^{1/k}\le1/R\), and the Taylor series of \(f\) at \(p\) converges on \(D_R(p)\), where it equals \(f\).
Differentiating the Cauchy integral formula \(f(w)=\frac{1}{2\pi i}\int_{\abs{z-p}=R}\frac{f(z)}{z-w}\dz\) under the integral sign \(k\) times at \(w=p\) gives the first two equalities. Parametrizing the circle by \(z=p+Re^{i\theta}\), so that \(\dz=iRe^{i\theta}\dtheta\), gives the third. Bounding the integrand of the third expression by \(M_R/R^k\) over an interval of length \(2\pi\) gives the estimate, and the root test gives the radius of convergence.