Proving the existence of point $z$ s.t. $f^{(n)}(z) = 0$ 
Suppose that $f$ is $n$ times differentiable on an interval $I$ and there are $n + 1$ points $x_0, x_1, \ldots, x_n \in I, x_0 < x_1 < \cdots < x_n$, such that $f(x_0) = f(x_1) = \cdots = f(x_n) = 0$. Prove that there exists a point $z \in I$ such that $f^{(n)}(z) = 0$.

I am trying to solve this, but other than using then using Rolle's Theorem, I am not sure how to proceed.

Here’s a fairly broad hint:
You know from Rolle’s theorem that it’s true when $n=1$. Try it for $n=2$; Rolle’s theorem gives you points $y_0$ and $y_1$ such that $x_0<y_0<x_1<y_1<x_2$ and $f\;'(y_0)=f\;'(y_1)=0$. Can you now apply Rolle’s theorem to $f\;'$ on some interval to get something useful?
In order to generalize the hint from $2$ to $n$, you’ll need to use mathematical induction.
