If $f \in \mathbf{C}^n(\mathbf{R})$ and $f(x_{0})=...=f(x_{n}) = 0$ for $x_{0} < ... < x_{n}$ then $f^{n}(\xi)=0$ for some $\xi \in (x_{0},x_{n})$ I was wondering if someone could look at my proof and tell me if this is sufficient enough.
Because $f$ is continuous on $\mathbf{R}$ and $f(x_{0})=f(x_{1})=0$ for $x_{0} < x_{1}$, we can apply rolle's theorem to show that $f'(a)=0$ for some $a \in (x_{0},x_{1})$. Similarly, we can apply rolle's theorem again since $f$ is continuous on $\mathbf{R}$ and $f(x_{1}) = f(x_{2})$ for $x_{1} < x_{2}$ to show that there exists a $f'(b)$ for some $b \in (x_{1},x_{2})$. Since $f'(a)=0$ for some $a \in (x_{0},x_{1})$ and $f'(b) = 0$ for some $b \in (x_{1},x_{2})$, we have $f''(c)=0$ for some $c \in (x_{0},x_{2})$ by rolle's theorem again. By applying rolle's theorem continuously n times, we have $f^{n}(\xi) = 0$ for some $\xi \in (x_{0},x_{n})$.
Thank you!
 A: You are using the correct tool. What remains is to write the answer in a manner that cannot be faulted by your enemy. Ideally, you try to make it clear to the reader what you are about to do and then you do it. Afterwards you should draw conclusions. There is no shame in writing in a mechanical style. The use of fixed terms and phrases allows the reader to concentrate on the mathematics.

In general, if $g : \mathbb{R} \rightarrow \mathbb{R}$ is differentiable and has at least $n+1$ distinct zeros, then $g'$ has at least $n$ distinct zeros that are interlaced with the zeros of $g$ much like the teeth of zipper.
Now let $a = \min x_j$ and let $b = \max x_j$. Let $S_n = \{1,2,3,\dotsc,n\} \subset \mathbb{N}$ and let $V \subseteq S_n$ be given by
$$ V = \{ k \in S_n \: | \: \text{$f^{(k)\:}$ has at least $n+1-k$ distinct zeros in $[a,b]$} \}.$$
We will now show that $V = S_n$ using the well-ordering principle. By assumption, $f$ has $n+1$ distinct zeros in $[a,b]$. By Rolle's theorem it follows that $f'$ has at least $n$ distinct zeros in $[a,b]$. This shows that $1 \in V$. Now assume that $V \not = S_n$. It follows that $S_n \setminus V \subseteq \mathbb{N}$ is not the empty set. Therefore, the well-ordering principle implies that $S_n \setminus V$ has a smallest element $k$. Since $1 \in V$, we must have $2 \leq k$. Now consider $k-1$. We must have $k-1 \in V$ or $k$ is not the smallest element of $S_n \setminus V$. This shows that $f^{(k-1)}$ has at least $n+2-k$ distinct zeros in $[a,b]$. By Rolle's theorem, $f^{(k)}$ has at least $n+1-k$ distinct zeros in $[a,b]$. This shows that $k \in V$. This is a contradiction because $k \in S_n \setminus V$. We conclude that $V = S_n$. In particular, $f^{(n)}$ has at least one zero in $[a,b]$ because $n \in V$. This completes the proof.
