**Proof by induction on $n$.** For $n = 0$: $p(x) = f[x_0] = f(x_0)$, which is the constant polynomial interpolating $f$ at $x_0$. $\checkmark$

**Inductive step.** Assume the result holds for $n-1$ points. Let $q \in P_{n-1}[x]$ be the polynomial interpolating $f$ at $x_0, \dots, x_{n-1}$, which by the inductive hypothesis has the Newton form $q(x) = f[x_0] + \sum_{k=1}^{n-1} f[x_0, \dots, x_k] \prod_{i=0}^{k-1}(x - x_i)$. Define \begin{align*} p(x) = q(x) + c \prod_{i=0}^{n-1}(x - x_i), \end{align*} where $c$ is chosen so that $p(x_n) = f(x_n)$. Since $q(x_i) = f(x_i)$ for $i = 0, \dots, n-1$, and $\prod_{i=0}^{n-1}(x_i - x_i) = 0$ for each such $i$, the polynomial $p$ interpolates $f$ at $x_0, \dots, x_{n-1}$ regardless of $c$. The condition $p(x_n) = f(x_n)$ gives \begin{align*} c = \frac{f(x_n) - q(x_n)}{\prod_{i=0}^{n-1}(x_n - x_i)}. \end{align*} It remains to show $c = f[x_0, \dots, x_n]$. By uniqueness of the interpolating polynomial ([Existence and Uniqueness](/theorems/473)), $p$ is the unique element of $P_n[x]$ interpolating $f$ at $x_0, \dots, x_n$. The leading coefficient of $p$ is $c$ (since $q$ has degree $\leq n-1$ and $\prod_{i=0}^{n-1}(x - x_i)$ is monic of degree $n$). But the leading coefficient of the degree-$n$ interpolant is $f[x_0, \dots, x_n]$ (this follows from the definition of divided differences: $f[x_0, \dots, x_n]$ is the coefficient of $x^n$ in the interpolant, as can be verified by expanding the Lagrange form). Therefore $c = f[x_0, \dots, x_n]$. $\blacksquare$