Prove that $f$ has a power series expansion around any point in its disc of convergence ${\bf Exercise:}$ Let $f$ be a power series centered at the origin. Prove that $f$ has a power series expansion around any point in its disc of convergence.
Proof:
We are given that $f = \sum_{n \geq 0} a_n z^n $ for $a_n \in \mathbb{C}$ and say it converges for values $z \in \mathbb{C}$  such that $|z| < R$. Take $z_0$ inside this disc arbitrary. Now, observe that
$$ z^n = (z-z_0 + z_0)^n = \sum_{k =0}^n {n \choose k} (z-z_0)^k z_0^{n-k} = (z-z_0)^n \sum_{k=0}^n {n \choose k} (z-z_0)^{-k} z_0^k$$
Now, write
$$ f(z) = \sum_{n \geq 0} a_n (z-z_0)^n \sum_{k=0}^n {n \choose k} (z-z_0)^{-k} z_0^k = \sum_{n \geq 0} c_n (z-z_0)^n$$
where $c_n = a_n \sum_{k=0}^n (z-z_0)^{-k} z_0^k $.
Is this good enough of an argument?
 A: There is a transformation error in the question. Instead of 
$$f(z) = \sum_{n \geq 0} a_n (z-z_0)^n \sum_{k=0}^n {n \choose k} (z-z_0)^{-k} z_0^k = \sum_{n \geq 0} c_n (z-z_0)^n,$$
with $c_n = a_n \sum_{k=0}^n (z-z_0)^{-k} z_0^k$, 
it should have been
$$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty c_n(z-z_0)^n,$$
where $c_n=\sum_{s=n}^\infty a_s\binom{s}{n} z_0^{s-n}$
More explicitly, as a formal computation,
$$\begin{align}
f(z)&=\sum_{n=0}^\infty a_nz^n= \displaystyle\sum_{n=0}^\infty a_n(z_0+(z-z_0))^n\\
&=\sum_{n=0}^\infty\sum_{k=0}^n a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\\
(\text{Switching the order of summation})\quad
&=\sum_{k=0}^\infty\sum_{n=k}^\infty a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\\
(\text{Replace $n$ by $s$ and $k$ by $n$ simultaneously})\quad &=\sum_{n=0}^\infty\left(\sum_{s=n}^\infty a_s\binom{s}{n} z_0^{s-n}\right)(z-z_0)^{n}.\\
\end{align}$$

If we substitute a complex number near $z_0$ for $z$ in the formal computation above, do all infinite sums thereof converge? Do all equalities hold? This is the question asked.
The answer is yes.
The basic idea is that absolute convergence implies convergence and, furthemore, the order of summation does not matter. That general principle can be stated in very general ways. What we need here is the following specific manifestation. 
If $d(k,n)\in \mathbb C$ for all integer $0\le k\le n$ and $\sum_{n=0}^\infty\sum_{k=0}^n \lVert d(k,n)\rVert$ converges, then we have the following.


*

*$\sum_{n=0}^\infty\left(\sum_{k=0}^n d(k,n)\right)$ converges

*$\sum_{n=k}^\infty d(k,n)$ converges for every fixed $k$, and $\sum_{k=0}^\infty\left(\sum_{n=k}^\infty d(k,n)\right)$ converges.

*$\sum_{n=0}^\infty\left(\sum_{k=0}^n d(k,n)\right)=
   \sum_{k=0}^\infty\left(\sum_{n=k}^\infty d(k,n)\right)$.


(If you are not familiar with absolute convergence, it should be a good exercise to prove the above.)

All we need to check now is the convergence of
$$\sum_{n=0}^\infty\sum_{k=0}^n \left\lVert a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\right\rVert.$$ 
Here we go. Suppose $\rVert z-z_0\rVert\lt R-\lVert z_0\rVert$. Then $\lVert z_0\rVert+ \lVert(z-z_0)\rVert\lt R$. We have
$$\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n \left\lVert a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\right\rVert &= \sum_{n=0}^\infty\sum_{k=0}^n \lVert a_n\rVert\binom{n}{k} \lVert z_0\rVert^{n-k}\lVert(z-z_0)\rVert^k
\\
&= \sum_{n=0}^\infty\lVert a_n\rVert(\lVert z_0\rVert+ \lVert(z-z_0)\rVert)^n\\
\end{align}$$
Since $R$ is the radius of convergence of $\sum_{n=0}^\infty a_nz^n$, the right-hand size converges. $\blacksquare$
