I have heard this few times, but no one had a proof of this, and I haven't succeeded to prove it my self.
I can prove it for $f:\mathbb{C}\to \mathbb{C}$, the proofs is as follows:
We can write $\mathbb{C}$ as follows.
$$ \mathbb{C} = \bigcup_{n=0}^{\infty} \left(f^{(n)}\right)^{-1} \left(\{0\}\right) $$
From Bair Category Theorem, so one of the $(f^{(n)})^{-1} (\{0\})$ contains open, denote this $n$ by $k$.
We know that $f^{(k)}$ is also smooth and that means analytic (since we are in $\mathbb{C}$) so $f^{(k)} \equiv 0$ so $f$ is polynomial.
I can also extend this argument for real analytic functions. We write $$ \mathbb{R} = \bigcup_{n=0}^{\infty} \left(f^{(n)}\right)^{-1} \left(\{0\}\right) $$
so there is $k$ such $(f^{(k)})^{-1} (\{0\})$ that contains open set. Using Zorn's Lemma, let $(a,b)$ be the maximal (w.r.t inclusion) open interval such that $(a,b)\subseteq (f^{(k)})^{-1} (\{0\})$ ($a$ and $b$ might be $\pm \infty$) We assume, w.l.o.g that $a$ is not $-\infty$. $f^{(k)}$ is continues so $f^{(k)}(a)=0$. Every derivative of $f^{(k)}$ is zero on $[a,b) = 0$ (on $(a,b)$ its clear. At $a$ we get then the one sided derivative is zero, but the derivative exists so its zero).
Looking at the taylor series at $a$, we see that its zero. We assumed that $ f$ is analytic, so there is positive converges radius around any point so there is a neighbourhood of $a$ witch the taylor series converges uniformly, so the taylor series of $f^{(k)}$ converges in that neighbourhood. This means that $f^{(k)}$ is zero on $(a-\epsilon,b)$ for some $\epsilon > 0 $, contradicting the maximality of $(a,b)$.
My question is: Does this statement still true for smooth (not necessarily analytic) real function? If not, do you have counter example?