[proofplan] We construct an auxiliary function $g$ that vanishes at $n+2$ distinct points (the $n+1$ interpolation nodes plus $x$), then apply Rolle's theorem repeatedly to find a zero of $g^{(n+1)}$. Evaluating $g^{(n+1)}$ using the facts that $p_n^{(n+1)} = 0$ (since $\deg p_n \le n$) and $\omega^{(n+1)} = (n+1)!$ (since $\omega$ is monic of degree $n+1$) yields the error formula. [/proofplan] [step:Reduce to the case $x \notin \{x_0, \dots, x_n\}$] If $x = x_j$ for some $j$, both sides of the error formula vanish ($f(x_j) - p_n(x_j) = 0$ and $\prod_{i=0}^n(x_j - x_i) = 0$), so any $\xi_x \in (a,b)$ works. Assume henceforth that $x \notin \{x_0, \dots, x_n\}$. [/step] [step:Construct an auxiliary function $g$ with $n+2$ zeros] Define the nodal polynomial $\omega(t) = \prod_{i=0}^n(t - x_i)$ and the auxiliary function \begin{align*} g(t) &= f(t) - p_n(t) - \frac{f(x) - p_n(x)}{\omega(x)}\,\omega(t). \end{align*} Then $g \in C^{n+1}[a,b]$, and $g$ vanishes at the $n+2$ distinct points $x_0, x_1, \dots, x_n, x$: at each $x_i$, $f(x_i) - p_n(x_i) = 0$ and $\omega(x_i) = 0$; at $x$, the construction forces $g(x) = f(x) - p_n(x) - \frac{f(x) - p_n(x)}{\omega(x)}\omega(x) = 0$. [/step] [step:Apply Rolle's theorem repeatedly to locate a zero of $g^{(n+1)}$] By Rolle's theorem applied repeatedly: $g$ has $n+2$ zeros, so $g'$ has at least $n+1$ zeros (one between each consecutive pair), $g''$ has at least $n$ zeros, and continuing, $g^{(n+1)}$ has at least one zero $\xi_x \in (a,b)$. [/step] [step:Evaluate $g^{(n+1)}(\xi_x)$ to derive the error formula] Since $p_n$ has degree $\le n$, $p_n^{(n+1)} = 0$. Since $\omega$ is monic of degree $n+1$, $\omega^{(n+1)} = (n+1)!$. Therefore \begin{align*} 0 = g^{(n+1)}(\xi_x) &= f^{(n+1)}(\xi_x) - 0 - \frac{f(x) - p_n(x)}{\omega(x)} \cdot (n+1)!. \end{align*} Solving for $f(x) - p_n(x)$: \begin{align*} f(x) - p_n(x) &= \frac{1}{(n+1)!}\,f^{(n+1)}(\xi_x)\,\omega(x) = \frac{1}{(n+1)!}\,f^{(n+1)}(\xi_x)\prod_{i=0}^n(x - x_i). \end{align*} [/step]