The following task is given:
Let $n \in \mathbb{N}$ and $f$ be a $n-$times differentiable function. With $f^{(n)}$ we name the nth derivation of $f$.
Show: If there exist $n+1$ different numbers $x_1 < ... < x_{n+1}$ with
$$f(x_j) = 0, \; \; j = 1, ..., n+1,$$
there exists $\xi \in (x_1, x_{n+1})$ with $f^{(n)}(\xi) = 0$.
I think this could easily be proved by induction. So the base clause $n = 1$ should be clear, as it is the application of Rolle's theorem on the 1-time differentiable function $f$.
So let's assume that the proposition holds for an $n \in \mathbb{N}$.
Let's further assume that $f$ is now $(n+1)-times$ differentiable and that there exist n+2 different numbers $x_1 < ... < x_{n+1} < x_{n+2}$, such that $$f(x_j) = 0, \; \; j = 1, ..., n+2$$
Now by induction hypothesis, there exist $\xi_1 \in (x_1, x_{n+1})$ such that $f^{(n)}(\xi_1) = 0$ and $\xi_2 \in (x_2, x_{n+2})$ such that $f^{(n)}(\xi_2) = 0$ and using the theorem of Rolle again, we should be done in case, that $\xi_1 \neq \xi_2$.
However my concept is pretty much over in case that $\xi_1 = \xi_2$. What could be done to fix this issue?