[proofplan] We expand the Cauchy kernel $1/(w - z)$ as a geometric series in $(z - a)/(w - a)$ and interchange summation with integration. The geometric series converges uniformly for $w$ on the circle $|w - a| = \rho$ when $|z - a| < \rho$, justifying the interchange. The resulting coefficients are identified with $f^{(n)}(a)/n!$ via the [Cauchy integral formula](/theorems/345) for derivatives. [/proofplan] [step:Apply the Cauchy integral formula and expand the kernel as a geometric series] Fix $z \in B(a, R)$ and choose $\rho$ with $|z - a| < \rho < R$. By the [Cauchy integral formula](/theorems/345) applied on the circle $|w - a| = \rho$: \begin{align*} f(z) = \frac{1}{2\pi i} \int_{|w - a| = \rho} \frac{f(w)}{w - z} \, dw. \end{align*} Rewrite the Cauchy kernel by factoring $w - a$ from the denominator: \begin{align*} \frac{1}{w - z} = \frac{1}{(w - a) - (z - a)} = \frac{1}{w - a} \cdot \frac{1}{1 - \frac{z - a}{w - a}}. \end{align*} Since $|z - a| < \rho = |w - a|$, the ratio $|(z-a)/(w-a)| = |z - a|/\rho < 1$, so the geometric series converges: \begin{align*} \frac{1}{w - z} = \sum_{n=0}^\infty \frac{(z - a)^n}{(w - a)^{n+1}}. \end{align*} [guided] The geometric series $\frac{1}{1 - u} = \sum_{n=0}^\infty u^n$ converges absolutely for $|u| < 1$. Here $u = (z-a)/(w-a)$, and for $w$ on the circle $|w-a| = \rho$ with $|z-a| < \rho$, we have $|u| = |z-a|/\rho < 1$. This is where the choice $|z-a| < \rho$ is essential: it guarantees the series converges. Moreover, the convergence is uniform in $w$ on $|w-a| = \rho$. The ratio $|u| = |z-a|/\rho$ is a constant independent of $w$ (it depends only on $z$ and $\rho$). Setting $q = |z-a|/\rho < 1$, the Weierstrass M-test applies with $M_n = q^n$, since $\sum q^n < \infty$. This uniform convergence is what justifies interchanging summation and integration in the next step. [/guided] [/step] [step:Interchange sum and integral to obtain the Taylor series] The convergence of the geometric series is uniform in $w$ on $|w - a| = \rho$ (since $|z-a|/\rho < 1$ is independent of $w$). We may therefore interchange sum and integral: \begin{align*} f(z) &= \frac{1}{2\pi i} \int_{|w-a|=\rho} \sum_{n=0}^\infty \frac{f(w)}{(w-a)^{n+1}} (z-a)^n \, dw \\ &= \sum_{n=0}^\infty \left( \frac{1}{2\pi i} \int_{|w-a|=\rho} \frac{f(w)}{(w-a)^{n+1}} \, dw \right) (z-a)^n = \sum_{n=0}^\infty c_n (z-a)^n, \end{align*} where \begin{align*} c_n = \frac{1}{2\pi i} \int_{|w-a|=\rho} \frac{f(w)}{(w-a)^{n+1}} \, dw. \end{align*} By the [Cauchy integral formula](/theorems/345) for derivatives, $c_n = f^{(n)}(a)/n!$. The series converges for all $z \in B(a, R)$ since $\rho \in (|z-a|, R)$ was arbitrary: for any $z \in B(a, R)$, choose $\rho \in (|z-a|, R)$ to obtain convergence at $z$. [/step]