Given $f(z)=\sum_{n=0}^{\infty} c_nz^n$ for all $|z|<R$, we want to show that for any $a$ such that $|a|<R$, $f(z)$ can be expanded as a power series centered at $a$ so that $f(z)=\sum_{n=0}^{\infty} d_n(z-a)^n$ for all $|z-a|<R-|a|$.
My book provides a proof, but it's somehow involved. I was wondering if there was a simple proof of the above result.
You first use binomial expansion and re-ordering, using absolute convergence inside the radius of convergence to justify the re-ordering, to get $$ \sum_{k\ge 0} c_kz^k=\sum_k c_k(a+(z-a))^k=\sum_{0\le j\le k}c_k\binom{k}{j}a^{k-j}(z-a)^j =\sum_{j\ge 0}\left(\sum_{k\ge j}c_k\binom{k}{j}a^{k-j}\right)(z-a)^j $$
Then use that by Cauchy-Hadamard (in the backward direction) $|c_k|\le br^{-k}$ for $r<R$ and some $b>0$ to conclude \begin{align} \left|\sum_{k\ge 0}c_{k+j}\binom{k+j}{j}a^{k} \right| &\le\sum_{k\ge 0}|c_{k+j}|\binom{k+j}{k}|a|^{k} \le br^{-j}\sum_{k\ge 0}\binom{-j-1}{k}\left(-r^{-1}|a|\right)^k\\[1em] &=br^{-j}(1-r^{-1}|a|)^{-1-j}=br(r-|a|)^{-1-j} \end{align} giving convergence by the root test resp. again Cauchy-Hadamard (now in the forward direction) for $|z-a|<r-|a|$. As this holds for all $r<R$, you get $R-|a|$ as radius of convergence for the shifted series.
I do not think that one can avoid these constructs, as you somehow have to address the coefficients and their asymptotic behavior.