Let $q_0 \in P_{k-j-1}[x]$ interpolate $f$ at $x_j, \dots, x_{k-1}$, let $q_1 \in P_{k-j-1}[x]$ interpolate $f$ at $x_{j+1}, \dots, x_k$, and let $q_2 \in P_{k-j}[x]$ interpolate $f$ at $x_j, \dots, x_k$. Define \begin{align*} r(x) := \frac{x - x_j}{x_k - x_j}\, q_1(x) + \frac{x_k - x}{x_k - x_j}\, q_0(x). \end{align*} This is a polynomial of degree at most $k - j$. For $i = j+1, \dots, k-1$, both $q_0$ and $q_1$ interpolate $f$ at $x_i$, so $r(x_i) = f(x_i)$. At $x_j$: the first term vanishes and $r(x_j) = q_0(x_j) = f(x_j)$. At $x_k$: the second term vanishes and $r(x_k) = q_1(x_k) = f(x_k)$. So $r$ interpolates $f$ at all of $x_j, \dots, x_k$. By the [uniqueness of interpolating polynomials](/theorems/473), $r = q_2$. Now $f[x_j, \dots, x_k]$, $f[x_{j+1}, \dots, x_k]$, and $f[x_j, \dots, x_{k-1}]$ are the leading coefficients of $q_2$, $q_1$, $q_0$ respectively. The leading coefficient of $r$ is $\frac{1}{x_k - x_j}$ times the leading coefficient of $q_1$ minus $\frac{1}{x_k - x_j}$ times the leading coefficient of $q_0$ (since $\frac{x - x_j}{x_k - x_j} q_1 - \frac{x_k - x}{x_k - x_j} q_0 = \frac{x}{x_k - x_j}(q_1 - q_0) + \text{lower order}$, and the leading terms of $q_1$ and $q_0$ contribute $\frac{f[x_{j+1}, \dots, x_k] - f[x_j, \dots, x_{k-1}]}{x_k - x_j}$). Equating with $f[x_j, \dots, x_k]$ gives the result. $\blacksquare$