Power series are analytic on their disk of convergence

Let $f(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$ be a power series with radius of convergence $R>0$.

Corollary: For every $x$ with $|x-x_0|<R$, the function $f$ is $C^\infty$ (infinitely differentiable ) and for each $k\ge 1$, $f^{(k)}(x)=\sum_{n=k}^\infty n(n-1)\cdots(n-k+1)\,a_n (x-x_0)^{n-k}.$ In particular, the Taylor series of $f$ at $x_0$ is exactly the original power series: $f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \quad (|x-x_0|<R).$

Connection to parent theorem: This follows by repeated application of the term-by-term differentiation theorem for power series, which preserves the radius of convergence and gives uniform convergence of derivatives on closed balls $|x-x_0|\le r<R$.