Let us define for all $z \in \mathbb{C}-\{0\} $ $$P_n(z)= 1+\frac{1}{z}+ \frac{1}{2!z^2}+\frac{1}{3!z^3}+ ...+\frac{1}{n!z^n}$$. Prove that $\forall {\epsilon>0}$ there exists $N$ such that $\forall n>N$ the following holds $$P_n(z)=0 \implies |z|<\epsilon$$
I would appreciate any hints or solutions to this problem.
Suppose this is not true. Then there exists $\epsilon >0$ such that for each $N$ there exist $k_N >N$ and a point $z_{k_N}$ with $p_{k_N}(z_{k_N}) =0$ and $|z_{k_N}| \geq \epsilon$. Since $\frac 1 {z_{k_N}} $ is bounded it has a convergent subsequence. Let $z$ be the limit of such a subsequence. By taking limit in $p_{k_N}(z_{k_N}) =0$ conclude that $\sum \frac {z^{n}} {n!}=0$ which means $e^{z}=0$. This is a contradiction.