= Solution
{c}
Use the endpoint <continuity> of $f$ from part <6/b>, and interpret the displayed endpoint <derivatives> as one-sided endpoint values. Starting with $f(a)=f(b)=0$, <Rolle's theorem> gives $c_1\in(a,b)$ with $f'(c_1)=0$. Since also $f'(b)=0$, apply <Rolle's theorem> to $f'$ on $[c_1,b]$ to obtain $c_2\in(c_1,b)$ with $f''(c_2)=0$. Repeating gives
$$
a<c_1<\cdots<c_{n+1}<b,\qquad f^{(j)}(c_j)=0\quad(1\leq j\leq n+1).
$$
For $j<n$, the lower <derivative> $f^{(j)}$ is <continuous> in the interior by <differentiability implies continuity>, and at $b$ by the existence of its next one-sided <derivative>. At the final step, the stipulated <continuity> of $f^{(n)}$ on $[a,b]$ and its <differentiability> on $(a,b)$ give exactly the hypotheses of <Rolle's theorem> on $[c_n,b]$. Therefore
$$
\boxed{f^{(n+1)}(c_{n+1})=0.}
$$
The endpoint regularity matters: merely assigning arbitrary values to $f$ at $a,b$ while prescribing only the interior <derivatives> would not justify the first application of <Rolle's theorem>.
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