 Solution

For distinct interpolation nodes $x_0,\dots ,x_n$, let $p_n$ be the unique polynomial of degree at most $n$ satisfying $p_n(x_i)=f(x_i)$. Existence follows from the Lagrange polynomial expressionuniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.

Define the divided difference $f[x_0,\dots ,x_n]$ to be the coefficient of $x^n$ in $p_n$, even when that coefficient is zero. For $n=0$ this gives $f[x_0]=f(x_0)$. Permuting the nodes does not change the interpolation conditions, so it does not change their unique interpolant or this coefficient. The divided difference is symmetric in all the nodes. Equivalently its symmetric expression is