= Solution
{c}
Let $\omega(t)=\prod_{r=0}^n(t-x_r)$. For $x$ not one of the nodes, put $c=(u(x)-p_n(x))/\omega(x)$ and define
$$
F(t)=u(t)-p_n(t)-c\omega(t).
$$
This function has zeros at the $n+1$ <interpolation> nodes and at the additional point $x$. Repeated <Rolle's theorem> gives a point $\eta$ between the outermost of these points such that $F^{(n+1)}(\eta)=0$. Since $p_n^{(n+1)}=0$ and $\omega^{(n+1)}=(n+1)!$,
$$
c=\frac{u^{(n+1)}(\eta)}{(n+1)!}.
$$
Substitution yields
$$
\boxed{u(x)-p_n(x)=
\frac{u^{(n+1)}(\eta)}{(n+1)!}\prod_{r=0}^n(x-x_r).}
$$
At an <interpolation> node both sides are zero, so any $\eta\in[-1,1]$ gives the identity there. The assumed $C^{n+1}$ regularity permits all applications of <Rolle's theorem>.
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