A function is differentiable $n$ times. Assume there are $n+1$ distinct points. Prove that $\exists$ one point $y$ such that $f^{(n)}(y)=0$ So I'm stuck on this question, I have an idea on the question but I missed the lecture which it pertained to. So I'm unsure of the theory behind it so, it'd be appreciated if someone could help me out! :)
Question
Suppose the function $f : \mathbb{R}\rightarrow\mathbb{R}$ is $n$ times differentiable on $\mathbb{R}$. Assume there are $n+1$ distinct points {$x_1, x_2,...,x_n,x_{n+1}$} such that $x_1<x_2<...<x_n<x_{n+1}$ and $f(x_i)=0$ for all $i=1,2,...,n,n+1$. Prove that there exists at least one point $y$ such that $f^{(n)}(y)=0$.
Note
Unfortunately I don't really have an attempt as I've been sitting on it for 2 hours now unsure where to even start, because as I said I missed the lecture, however I have come to the conclusion that it could possibly involve doing Rolle's theorem multiple times but I don't really know how to actually apply it, etc. Anyway, ANY help would GREATLY be appreciated! :)

Yes, you are going right way, applying Rolle's Theorem to the interval $(x_1,x_{n+1})$ we get that $f^{(1)}(x)$ must have at least $n$ roots in the interval $(x_1,x_{n+1})$ 
Now again apply Rolle's Theorem for the function  $f^{(1)}(x)$ from which you get that  $f^{(2)}(x)$ must have at least $n-1$ roots in the interval $(x_1,x_{n+1})$ .
Doing so you'll get that  $f^{(i)}(x)$ must have at least $n+1-i$ roots in the interval $(x_1,x_{n+1})$ .
Now for $i=n$ you'll have at least $n+1-n$ root, I.e. at least $1$ root in the given interval .
For applying Rolle's Theorem you need : 

For interval $[a,b]$ if $f$ is continuous and differentiable, and $f(a) =f(b)$ then $f'(x)$ must be zero for at least on $x$ in the interval $[a,b]$

