Derivative polynomial Let $f :\mathbb R\to\mathbb R$ be an infinitely differentiable  function. Assume that
for every $x \in \mathbb R$, there exists an  $n_x\in \mathbb N$, such that $\,f^{(n_x)}(x)=0$. Prove that  $f$ is polynomial.
 A: This is definitely correct if we assume that the function $f$ is real analytic. In such case $f$ extends holomorphically in an open neighbourhood $\Omega$ of the real line, so we can imagine $f$ as a holomorphic function in $\Omega$, with $\mathbb R\subset \Omega$.
Let $S_n=\{x\in\mathbb R: f^{(n)}(x)=0 \}$. Then $\bigcup_{n\in\mathbb N} S_n=\mathbb R$. Clearly at least one of the $S_n$'s is uncountable, and thus it has an accumulation point $x_0$. That means, that the zero's of $f^{(n)}$ have an accumulation point, which implies that $f^{(n)}$ is identically zero, and thus $f$ a polynomial of degree at most $n-1$. 
In the $C^\infty$ case now, I have not produced a proof yet, but this is as much as I could get.
As the $S_n$'s are closed, Baire's Theorem implies the existence of set $S_n$ with nonempty interior. Thus there exists an interval $(a,b)$ where $f^{(n)}$ vanishes and thus $f|_{(a,b)}$ is a polynomial of degree at most $n-1$, and clearly the are such interval (for different values of $n$) dense in the whole real line. 
