= Solution
For distinct <interpolation nodes> $x_0,\ldots,x_n$, 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> expression
$$
p_n(x)=\sum_{i=0}^nf(x_i)\prod_{j\ne i}\frac{x-x_j}{x_i-x_j};
$$
uniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.
Define the <divided difference> $f[x_0,\ldots,x_n]$ 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. \b[The divided difference is symmetric in all the nodes.] Equivalently its symmetric expression is
$$
f[x_0,\ldots,x_n]=\sum_{i=0}^n\frac{f(x_i)}{\prod_{j\ne i}(x_i-x_j)}.
$$
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