Entire function with vanishing derivatives? Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function.
And assume that at each point, one of it's derivatives vanishes.
What can you say about $f$?
A hint suggests that $f$ must be a polynomial.
 A: Let $A_n$ be the set of points where $f^{(n)}(z)$ vanishes.
Each $A_n$ is closed. By the Baire category theorem one of those sets must have a non empty interior which in turn implies that some derivative vanishes on an open ball.
Therefore it is identically zero and $f$ is a polynomial.
A: Let $n: [0,1] \to \{0,1,... \}$ be defined such that $n(x)$ is the smallest $k \ge 0$ such that $f^{(k)}(x) = 0$.
Since $[0,1]$ is uncountable, there must be some $k$ such that $Z=n^{-1} \{k\}$ has an infinite number of points. Since $[0,1]$ is compact, $Z$ has a limit point $p \in [0,1]$, and hence we have $f^{(k)}(z) = 0$ everywhere.
Elaboration:
Let $p_n \in Z$ such that $p_n \to p$. Since $f^{(k)}(p_n) = 0$, the
identify theorem (https://en.wikipedia.org/wiki/Identity_theorem) shows that $f^{(k)}(z) = 0$ everywhere from which it follows that $f$ is a polynomial (take the power series expansion around any point).
A: we can also solve this without using baire category theorem..
we can  use the fact that zeros of non zero analytic  function is countable. Note that union of $A_{n}$ is  $\mathbb{C}$ thus we must have some fix derivative identically 0. Now use power series and analytic continuation.
