= Solution
If $x$ is a node the error and the product both vanish, and any interior $\xi$ works. Otherwise put $\omega(t)=\prod_{i=0}^n(t-x_i)$ and $C=(f(x)-p_n(x))/\omega(x)$. The <function>
$$
R(t)=f(t)-p_n(t)-C\omega(t)
$$
has $n+2$ distinct zeros, at the nodes and at $x$. Repeated <Rolle's theorem> gives a point $\xi$ between the extreme zeros, and therefore in $(a,b)$, where $R^{(n+1)}(\xi)=0$. Since $p_n^{(n+1)}=0$ and $\omega^{(n+1)}=(n+1)!$, one gets $C=f^{(n+1)}(\xi)/(n+1)!$. Substitution gives
$$
\boxed{f(x)-p_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i)}.
$$
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