Prove that $f$ is a polynomial if one of the coefficients in its Taylor expansion is 0 Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion
$$f(z) = \sum_{n = 0}^{\infty}c_n(z-z_0)^n$$
is equal to $0$. Prove that $f$ is a polynomial.
The problem hints to use the fact that $c_nn! = f^{(n)}(z_0)$ and use a countability argument.
My attempt at a solution
The only thing I can think of in this case is that the coefficients $c_n$ of the above Taylor expansion are defined as:
$$c_n = \frac{f^{(n)}(z_0)}{n!}$$
The only way one of these $c_n$ could be zero is if the derivative of order $n$ vanishes everywhere. Thus, this would mean that $f$ is a polynomial of order $k < n$.
Does this suffice as a proof? How would I use a countability argument to prove this instead? Thanks.
 A: If none of the derivatives are exactly zero, the set of $a \in \mathbb{C}$ such that there exists $n \in \mathbb{N}$ such that $f^{(n)}(a)=0$, i.e. $\cup_{n=0}^\infty (f^{(n)})^{-1}(0)$  is countable. This uses that an analytic function has only countably many zeroes, the quickest way I can see to prove this is the fact it has only finitely many in any compact set (or you get a limit point) and $\mathbb{C}$ is a countable union of compact sets (e.g. positive integer radius balls around 0)
A: No, that's not true.  It's quite possible for some derivative of $f$ to be $0$ at some points $a$ without vanishing everywhere.  What you have to use is that
for every $a$ there is such an $n$.  
Hints:


*

*The zeros of a non-constant analytic function form a discrete set, i.e. have no limit point in the domain of the function.

*A discrete subset of $\mathbb C$ is countable.


BTW: there is a much harder version of the problem that replaces $\mathbb C$ by a real interval and replaces analytic by $C^\infty$.
