Assume $f$ is an entire function such that for each $z\in \mathbb{C}$, we have an $n=n(z)$ such that $f^{(n)}(z)=0$, show that $f$ is a polynomial.
I got stuck in my following argument.
First, we only need to show that we have a uniform $n$ such that $f^{(n)}=0$.
To show this, we only need to show that we have a uniform $n$ such that $f^{(n)}|_{\bar{D}}=0$, where $\bar{D}=\{z||z|\leq 1\}$.
Suppose not, we get a sequence $z_n\in \bar{D}$, such that $f^{(n)}(z_n)\neq 0$, then W.L.O.G., we can assume $z_n\rightarrow z\in \bar{D}$, suppose $f^{(m)}(z)=0$, then ???