TITLE: prove that the entire function f is a polynomial.
QUESTION [8 upvotes]: Suppose that $f$ is an entire function, and that in every power series 
$f(z)=\sum_{n=0}^{\infty} c_{n}(z-a)^n$
at least one coefficient is 0. Prove that $f$ is a polnomial.
Hint: $n!c_{n}=f^{(n)}(a)$
Actually, this is a Rudin's book's exercise.
I tried Cauchy inequality, and Liouville's theorem ( for $g(z)=\sum_{n=m}^{\infty} c_{n}(z-a)^n$ is bounded) but failed.
I really want to solve that, but I don't have any idea. I need your help.

REPLY [11 votes]: From the hint, we see that for every $c \in \mathbf C$, some higher derivative of $f$ vanishes at $c$. Let $Z_n = \{z \in \mathbf C : f^{(n)}(z)=0\}$. If no derivative of $f$ vanishes identically, then each $Z_n$ is a closed, nowhere dense subset of $\mathbf C$. But $\mathbf C = \bigcup_{n\geq 0} Z_n$ is then impossible by the Baire Category Theorem. Thus some derivative of $f$ vanishes identically, which implies $f$ is a polynomial.