TITLE: Proving $(n+1)$th differential is $0$ given lower differentials are $0$
QUESTION [6 upvotes]: Following is a question I am stuck in.

Let $f : \Bbb R \to \Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
  $$f(1) = f(0) = f^{(1)}(0) = f^{(2)}(0) = · · · = f^{(n)}(0) = 0$$
  Prove that there exists $x \in (0, 1)$ such that $f^{(n+1)}(x) = 0$.

It is a past question of an entrance exam. 
I thought to use Rolle's Theorem, but this requires information about $f^{(n)} (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$

REPLY [4 votes]: Love the answers. Another method. 
Taylor expansion: 
$$f(x) = f(0)+ f'(0)x+\dots + f^{(n)}(0)\frac{x^n}{n!}+\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$ 
$c$ is in the open interval between $0$ and $x$.