[proofplan] We prove three facts in sequence. First, the Hadamard formula $1/R = \limsup_{n \to \infty} |c_n|^{1/n}$ shows the derived [series](/page/Series) $\sum n c_n (z-a)^{n-1}$ has the same [radius of convergence](/theorems/273) $R$, since $n^{1/n} \to 1$. Second, we show the difference quotient $(f(z) - f(w))/(z - w)$ converges to $g(w) = \sum n c_n (z-a)^{n-1}$ by factoring $(z-a)^n - (w-a)^n$ and extracting a factor of $(z-w)$. Third, induction gives infinite [differentiability](/page/Derivative) and the coefficient formula $c_n = f^{(n)}(a)/n!$. [/proofplan] [step:Show the derived series has the same radius of convergence] Let $g(z) = \sum_{n=1}^\infty n c_n (z-a)^{n-1}$. The [radius of convergence](/theorems/273) of $\sum n c_n (z-a)^{n-1}$ is determined by the Hadamard formula: \begin{align*} \frac{1}{R'} = \limsup_{n \to \infty} |nc_n|^{1/n}. \end{align*} Since $\lim_{n \to \infty} n^{1/n} = 1$, we have $|nc_n|^{1/n} = n^{1/n} |c_n|^{1/n}$, and therefore \begin{align*} \limsup_{n \to \infty} |nc_n|^{1/n} = 1 \cdot \limsup_{n \to \infty} |c_n|^{1/n} = \frac{1}{R}. \end{align*} Hence $R' = R$. [/step] [step:Prove $f'(w) = g(w)$ by estimating the difference quotient error] Fix $w \in B(a; R)$ and choose $r$ with $|w - a| < r < R$. For $z \neq w$ with $|z - a| < r$: \begin{align*} \frac{f(z) - f(w)}{z - w} - g(w) = \sum_{n=1}^\infty c_n \left(\frac{(z-a)^n - (w-a)^n}{z - w} - n(w-a)^{n-1}\right). \end{align*} Using the factorisation $(z-a)^n - (w-a)^n = (z-w)\sum_{k=0}^{n-1}(z-a)^k(w-a)^{n-1-k}$, the $n$-th term becomes \begin{align*} c_n \sum_{k=0}^{n-1}\bigl((z-a)^k - (w-a)^k\bigr)(w-a)^{n-1-k}. \end{align*} Applying the factorisation again to $(z-a)^k - (w-a)^k = (z-w)\sum_{j=0}^{k-1}(z-a)^j(w-a)^{k-1-j}$, each term acquires a factor of $(z-w)$. Bounding $|z-a|, |w-a| \leq r$: \begin{align*} \left|\frac{f(z) - f(w)}{z - w} - g(w)\right| \leq |z - w| \sum_{n=2}^\infty |c_n| \cdot \frac{n(n-1)}{2} r^{n-2}. \end{align*} The [series](/page/Series) $\sum |c_n| n(n-1) r^{n-2}$ converges (it is the second [derivative](/page/Derivative) of $\sum |c_n| t^n$ evaluated at $t = r < R$). Therefore the right side tends to $0$ as $z \to w$, giving $f'(w) = g(w)$. [guided] The double factorisation is the heart of the proof. We explain why two factorisations are needed and what each accomplishes. **First factorisation:** $(z-a)^n - (w-a)^n = (z-w)\sum_{k=0}^{n-1}(z-a)^k(w-a)^{n-1-k}$. This cancels the denominator $(z-w)$ in the difference quotient, but the resulting sum $\sum_{k=0}^{n-1}(z-a)^k(w-a)^{n-1-k}$ is not equal to $n(w-a)^{n-1}$ (which is what $g(w)$ contains). Their difference is $\sum_{k=0}^{n-1}((z-a)^k - (w-a)^k)(w-a)^{n-1-k}$. **Second factorisation:** Each term $(z-a)^k - (w-a)^k$ factors as $(z-w)\sum_{j=0}^{k-1}(z-a)^j(w-a)^{k-1-j}$. This extracts another factor of $(z-w)$, so the entire error acquires a factor of $|z-w|$. **Why the error vanishes:** The bound becomes $|z-w| \cdot \sum_{n=2}^\infty |c_n| \cdot \frac{n(n-1)}{2} r^{n-2}$. The series $\sum |c_n| n(n-1) r^{n-2}$ converges because it is the second derivative of $\sum |c_n| t^n$ evaluated at $t = r < R$ (and the [radius of convergence](/theorems/273) is unchanged by differentiation, as proved in the first step). As $z \to w$, the factor $|z - w| \to 0$ while the series remains bounded, so the entire expression tends to $0$, giving $f'(w) = g(w)$. [/guided] [/step] [step:Conclude infinite differentiability and the coefficient formula by induction] Since $f'(z) = g(z) = \sum_{n=1}^\infty n c_n (z-a)^{n-1}$ is itself a [power series](/page/Power%20Series) with radius $R$, the same argument applies to $f'$. By induction, $f$ is $k$-times [differentiable](/page/Derivative) for every $k \geq 1$, with \begin{align*} f^{(k)}(z) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1) c_n (z-a)^{n-k}. \end{align*} Evaluating at $z = a$: all terms with $n > k$ vanish (they contain the factor $(a-a)^{n-k} = 0$), leaving $f^{(k)}(a) = k! \, c_k$. Therefore $c_n = f^{(n)}(a)/n!$. [/step]