[proofplan]
We first iterate termwise differentiation for [power series](/page/Power%20Series). This gives an explicit formula for the $k$-th derivative of $f$ on the whole disk $B(z_0,R)$, and each differentiated series still has [radius of convergence](/theorems/262) $R$. Evaluating that formula at the center $z_0$ kills every term with positive power of $z-z_0$, leaving exactly the coefficient $k!a_k$. Dividing by $k!$ gives the desired coefficient formula.
[/proofplan]
[step:Iterate termwise differentiation to obtain the formula for every derivative]
For each pair of integers $k \ge 0$ and $m \ge k$, define the coefficient
\begin{align*}
c_{k,m}:=\frac{m!}{(m-k)!}a_m.
\end{align*}
For each integer $k \ge 0$, define the power series map $F_k:B(z_0,R)\to\mathbb{C}$ by
\begin{align*}
F_k(z):=\sum_{m=k}^{\infty} c_{k,m}(z-z_0)^{m-k}.
\end{align*}
We prove by induction on $k$ that $f$ is $k$ times complex differentiable on $B(z_0,R)$ and
\begin{align*}
f^{(k)}(z)=F_k(z)
\end{align*}
for every $z \in B(z_0,R)$.
For $k=0$, this is exactly the assumed power series representation of $f$, since $c_{0,m}=a_m$ for every integer $m\ge 0$.
Assume the assertion holds for some integer $k\ge 0$. Then
\begin{align*}
f^{(k)}(z)=\sum_{m=k}^{\infty}\frac{m!}{(m-k)!}a_m(z-z_0)^{m-k}
\end{align*}
for every $z\in B(z_0,R)$. This is a power series centered at $z_0$ with [radius of convergence](/theorems/265) at least $R$: for $k=0$ this is part of the hypothesis, and for $k\ge 1$ it follows inductively from the radius-preservation conclusion in the [Termwise Differentiation of a Power Series](/theorems/8249) theorem [citetheorem:8249]. Since $R>0$, this radius is positive, and the disk $B(z_0,R)$ lies inside the convergence disk on which that theorem gives termwise differentiation. Applying that theorem to the power series defining $f^{(k)}$ shows that $f^{(k)}$ is complex differentiable on $B(z_0,R)$ and
\begin{align*}
f^{(k+1)}(z)=\sum_{m=k+1}^{\infty}(m-k)\frac{m!}{(m-k)!}a_m(z-z_0)^{m-k-1}.
\end{align*}
Since
\begin{align*}
(m-k)\frac{m!}{(m-k)!}=\frac{m!}{(m-k-1)!},
\end{align*}
we obtain
\begin{align*}
f^{(k+1)}(z)=\sum_{m=k+1}^{\infty}\frac{m!}{(m-k-1)!}a_m(z-z_0)^{m-k-1}=F_{k+1}(z).
\end{align*}
Thus the induction is complete.
[guided]
The purpose of this step is to turn the informal phrase “differentiate the series repeatedly” into a precise induction. For each integer $k\ge 0$, we define the expected $k$-th derivative series as a map $F_k:B(z_0,R)\to\mathbb{C}$ by
\begin{align*}
F_k(z):=\sum_{m=k}^{\infty}\frac{m!}{(m-k)!}a_m(z-z_0)^{m-k}.
\end{align*}
The coefficient $\frac{m!}{(m-k)!}$ is the product obtained by differentiating $(z-z_0)^m$ exactly $k$ times.
We prove by induction on $k$ that $f^{(k)}=F_k$ on $B(z_0,R)$. For $k=0$, the formula says
\begin{align*}
F_0(z)=\sum_{m=0}^{\infty}a_m(z-z_0)^m,
\end{align*}
which is exactly the hypothesis that this power series represents $f$ on $B(z_0,R)$.
Now assume that the formula has been proved for some integer $k\ge 0$. Thus, for every $z\in B(z_0,R)$,
\begin{align*}
f^{(k)}(z)=\sum_{m=k}^{\infty}\frac{m!}{(m-k)!}a_m(z-z_0)^{m-k}.
\end{align*}
To differentiate this expression, we must justify termwise differentiation. The Termwise Differentiation of a Power Series theorem [citetheorem:8249] applies to a power series centered at $z_0$ with positive [radius of convergence](/theorems/273) and states that the differentiated series has the same radius of convergence. The series above has radius of convergence at least $R$: for $k=0$ this is part of the hypothesis on the original power series, and for $k\ge 1$ this lower bound for the radius is preserved at each previous differentiation by the same theorem. Because $R>0$, the relevant radius is positive, and because it is at least $R$, every point of $B(z_0,R)$ lies in the disk where the theorem gives termwise differentiation. Therefore [citetheorem:8249] applies to the power series defining $f^{(k)}$ on the disk $B(z_0,R)$.
Termwise differentiation gives
\begin{align*}
f^{(k+1)}(z)=\sum_{m=k+1}^{\infty}(m-k)\frac{m!}{(m-k)!}a_m(z-z_0)^{m-k-1}.
\end{align*}
The index starts at $m=k+1$ because the $m=k$ term is constant and differentiates to zero. Simplifying the coefficient gives
\begin{align*}
(m-k)\frac{m!}{(m-k)!}=\frac{m!}{(m-k-1)!}.
\end{align*}
Hence
\begin{align*}
f^{(k+1)}(z)=\sum_{m=k+1}^{\infty}\frac{m!}{(m-k-1)!}a_m(z-z_0)^{m-k-1}=F_{k+1}(z).
\end{align*}
This proves the induction step, so $f$ has complex derivatives of every order on $B(z_0,R)$ and the displayed formula holds for every derivative order.
[/guided]
[/step]
[step:Evaluate the derivative formula at the center of the disk]
Let $n\ge 0$ be an integer. Since $R>0$, the center $z_0$ belongs to $B(z_0,R)$. Applying the derivative formula with $k=n$ and $z=z_0$ gives
\begin{align*}
f^{(n)}(z_0)=\sum_{m=n}^{\infty}\frac{m!}{(m-n)!}a_m(z_0-z_0)^{m-n}.
\end{align*}
The term with $m=n$ is
\begin{align*}
\frac{n!}{0!}a_n(z_0-z_0)^0=n!a_n.
\end{align*}
Every term with $m>n$ contains the positive power $(z_0-z_0)^{m-n}=0^{m-n}=0$, so those terms vanish. Therefore
\begin{align*}
f^{(n)}(z_0)=n!a_n.
\end{align*}
Since $n! \neq 0$ in $\mathbb{C}$, division by $n!$ yields
\begin{align*}
a_n=\frac{f^{(n)}(z_0)}{n!}.
\end{align*}
This proves the coefficient formula for every integer $n\ge 0$.
[/step]
