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Given that $f$ is a holomorphic function on a domain $D \subset \mathbb{C}$, and $ \forall a \in D$ we have a $n_a \in \mathbb{N}$ such that $f^{n_a}(a) = 0$. We have to use this to show that $f$ is a polynomial in $z$.

My thoughts: As $f$ is holomorphic, it is analytic, and it has a power series expansion $\forall a \in D$. The derivative condition translates to the fact that in every power series expansion of $f$ around $a$, $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$, at least one coefficient $c_n$ is $0$.

Is my interpretation of the problem so far correct? How do I proceed from here? I'll probably have to use something like $\mathbb{N}$ is countable, but $\mathbb{C}$ is not. But still I'm not very clear.

The Limit Does Not Exist
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    this has been done previously here, but you can use that since $\mathbb N$ is countable there is an uncountable set of $a \in D_1$ subdomain st $\overline {D_1} \subset D$ which have the same $n_a=m$ so they have a limit point in $\bar D_1$ hence in $D$ and that implies $f^{(m)}=0$ in $D$ by the identity theorem – Conrad Feb 10 '23 at 05:05
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    Some more: https://math.stackexchange.com/questions/linked/995105?lq=1 – Martin R Feb 10 '23 at 05:36

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