Zero points of derivatives It's obvious that if $f(x)$ is a polynomial then it's derivatives $f^{(n)}$ are equal to zero 
for $n>\deg f$. 
I'm trying to prove the "inverse" statement: if for each $x\in\mathbb{R}$ there exist $n\in\mathbb{R}$ such that $f^{(n)}(x)=0$ then $f$ is a polynomial.
I've already proved using Baire theorem that there exist $[a,b]\subset\mathbb{R}$ and $i\in\mathbb{N}$ such that $f^{(i)}(x)=0$ for every $x\in[a,b]$, but I have no idea how to go further.
 A: This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Andrey Gogolev below.

The proof is by contradiction. Assume $f$ is not a polynomial.
Consider the following closed sets:
  $$
S_n = \{x: f^{(n)}(x) = 0\} 
$$
  and
  $$
X = \{x: \forall (a,b)\ni x: f\restriction_{(a,b)}\text{ is not a polynomial} \}. 
$$
It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and
  $$
(a,b)\cap X\subset S_n
$$
  for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S_m$ for all $m\ge n$ and $x\in (a,b)\cap X$. 
Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d< n$. (Since either $c$ or $e$ is in $X$.)
So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.

