Problem
Let $p(z)=(z-a_1)(z-a_2)…(z-a_N)$, where $a_1, a_2, …, a_N$ are distinct complex numbers. Let $M=\min_{1\le{k}\le{N}}|a_k|$. Prove that it is possible to express $\frac{1}{p(z)}$ as a power series $\sum_{n=0}^{\infty}c_nz^n$, for $|z|<M$.
Progress
We look to prove this by induction on $N$. The case $N=1$ is rather simple. We see that $$\frac{1}{p(z)}=\frac{1}{z-a_1}=-\frac{1}{a_1}\sum_{n=0}^{\infty}(\frac{z}{a_1})^n$$ which converges for $|z|<M$ by comparison with $\sum_{n=0}^{\infty}z^n$.
We assume now that the proposition holds for arbitrary $N$ and consider the '$N+1$' case.
Now, $p(z)=(z-a_1)(z-a_2)\cdots(z-a_N)(z-a_{N+1})$. By the assumption of our inductive hypothesis, $\frac{1}{(z-a_1)(z-a_2)\cdots(z-a_N)}$ can be expressed in the form $\sum_{n=0}^{\infty}c_nz^n$ for some complex coefficients $c_n$.
As such, $$\frac{1}{p(z)}=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}c_nz^n\sum_{n=0}^{\infty}\frac{1}{(a_{N+1})^n}z^n=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{c_n}{(a_{N+1})^{n-k}} z^n$$ which is of the correct form, but I'm not sure how to demonstrate that convergence holds for $|z|<M$
Any help would be very appreciated. I'm not sure if induction is the best approach to proving this.
EDIT 1
I'll leave previous working here for reference. It seems simply writing $\frac{1}{p(z)}$ as a linear combination of the fractions $\frac{1}{z-a_k}$ for $1\le{k}\le{N}$ is a far less cumbersome approach. Thanks to all who have helped.
Further Problem:
Could the radius of convergence exceed $M$?
Best Answer
Write $f = C \prod_i \frac{1}{1 - z/a_i}$, each has the regular power series expansion for $\frac{1}{1+z}$ for $|z| < |a_i|$, done!