Show that $f$ is a polynomial

Question

$f:\Bbb C\to \Bbb C$ be an entire function.If for each $z\in \Bbb C$ $\exists$ a positive integer $n$ such that $f^n(z)=0$ then show that $f$ is a polynomial

My try: Since $f$ is entire $f$ has a power series expansion about say a point $a$.

Then $f(z)=c_0+c_1(z-a)+...+c_n(z-a)^n+...\forall z$ where $c_n=\dfrac{f^n(a)}{n!}$

Then by given hypothesis for the given $a\in \Bbb C ;\exists m$ such that $c_m=\dfrac{f^m(a)}{m!}=0\implies f^m(a)=0$ and this holds forall $a\in \Bbb C$.

Since $\Bbb N$ is countable and $\Bbb C$ is uncountable so there exists at least one $n$ such that $f^n(a)=0$ for uncountable number of $a\in \Bbb C$. So an entire function can only have countable number of zeros so $f^n(z)\equiv 0$. So $f$ is a polynomial.

But I can't figure out if its correct or not.Please help me to figure it out.Is there any flaw in logic.

Answer 1

Hint:

Observe that $$\mathbb{C} = \bigcup_{n \in \mathbb{N}} M_n$$ with $$M_n := \{z \in \mathbb{C} \mid f^n(z) = 0\}.$$ What do you know about the cardinality of $M_n$?