Show that there exists $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$, one has $a_n = 0$, in other words show $f$ is a polynomial function. [duplicate]

Question

Let $\sum_{n=0}^{\infty}a_nz^n \in \mathbb{C}[[T]]$ be a power series over $\mathbb{C}$ with radius of convergence $R = \infty$, let $f: \mathbb{C} \to \mathbb{C}$ be the entire function given by $f(z) = \sum_{n=0}^{\infty}a_nz^n$. For every $z_0 \in \mathbb{C}$, let $$f(z) = \sum_{k=0}^\infty b_{k,z_0}(z-z_0)^k~~~\text{for } z \in \mathbb{C}$$ be the power series expansion of $f$ about $z_0$. Suppose for every $z_0 \in \mathbb{C}$, there exists $k \in \mathbb{N}$ (depending on $z_0$) such that $b_{k,z_0}=0$. Show that there exists $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$, one has $a_n = 0$, in other words show $f$ is a polynomial function.

I was trying to prove by contradiction that $f$ is not polynomial then $f(1/z)$ has essential singularity at $0$. But really have no idea how to continue... Any one has any idea.

Answer 1

For some $k$ $\{z:f^{(k)}(z)=0$ is uncountable (because $\mathbb C$ is uncountable ). This implies that this set has limit point, so $f^{(k)}\equiv 0$ and $f$ is a polynomial.