= Solution
{c}
Apply <Rolle's theorem> on each $[x_i,x_{i+1}]$. This gives $n$ ordered zeros of the <derivative> $f'$ strictly between the original $n+1$ zeros of $f$. Apply <Rolle's theorem> to $f'$ between each consecutive pair of its zeros to obtain $n-1$ ordered zeros of $f''$. Continue in this way.
To justify the iteration, for $j<n$, the <derivative> $f^{(j)}$ is <differentiable> throughout $(a,b)$ and hence <continuous> there by <differentiability implies continuity>. Every new closed <interval> lies strictly inside $(a,b)$, so <Rolle's theorem> applies at each step. After $j$ steps there are $n+1-j$ zeros of $f^{(j)}$. At $j=n$ this gives
$$
\boxed{\exists c\in(a,b):\quad f^{(n)}(c)=0.}
$$
This is the distinct-zero case of the <Rolle root count with multiplicities>.
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