(Proof) Analytic function have isolated zeros?

Question

A complex-polynomial of degree $n$ has $n$ roots, hence a complex power series can be thought of degree $\mid \mathbb{N} \mid$ and hence has countable many roots. I know my reasoning is not at all mathematically rigor, but can some prove the above question using same/similar argument.

Answer 1

Can't prove it because it's false. $e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$ has no roots.

That was an answer to the original version. It's hard to predict what the question's going to be when I finish this, but:

Yes, a non-trivial power series has at most countably many roots. This follows from the fact that the zeroes are isolated.

I suspect that the answer to what I suspect the question is is no, for at least two reasons: (i) II don't see how the fact that a polynomial has only $n$ roots implies directly that a power series has only countably many zeroes, and (ii) even if I did see how that goes I don't see how it would help: A countable set need not be isolated.