Proving $(n+1)$th differential is $0$ given lower differentials are $0$

Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read

If $f(1) = f(0) = f'(0) = 0$, there is $x \in (0,1)$ such that $f''(x) = 0$.

How would one go about this? Ok, if $f(1) = f(0)$ there is $y \in (0,1)$ such that $f'(y) = 0$. Now there is $x \in (0,y)\subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?

Since $f(0)=f(1)=0,\ \exists c\in(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0,\ \exists d \in (0,c)$ such that $f''(d)=0.$ Proceed.

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 $x$.