Divided differences are symmetric leading coefficients

For each $j$ define $L_j(t)={\prod }_{k\mathrm{\ne }j}(t-t_k)\mathrm{/}(t_j-t_k)$. Its denominators are nonzero, and $L_j(t_k)$ equals $1$ when $k=j$ and $0$ otherwise. Sort the nodes increasingly, carrying their data along. Reindexing the finite sum and products does not change them, so [F2] applies and $P(t)={\sum }_{j=0}^{q}f(t_j)L_j(t)$ is the unique polynomial of degree at most $q$ taking the assigned values.

The coefficient of $t^q$ in $P$ is the displayed sum, because the leading coefficient of $L_j$ is $1\mathrm{/}{\prod }_{k\mathrm{\ne }j}(t_j-t_k)$. The sum is unchanged by any permutation of the nodes, since the permutation only reorders its terms and their symmetric denominators.

We prove that the recursively defined divided difference equals this leading coefficient by induction on $q$ using [F4]. Since $0\in \mathbb{N}$ by [F3], for $q=0$ both are $f(t_0)$. For $q\ge 1$, substitute the induction formula for the two order-$(q-1)$ differences in [F1] and divide by $t_q-t_0$. For an interior index $0<j<q$, the resulting coefficient of $f(t_j)$ is $\frac{1}{t_q-t_0}(\frac{1}{t_j-t_q}-\frac{1}{t_j-t_0})\frac{1}{{\prod }_{\begin{array}{c}1\le k\le q-1\\ k\mathrm{\ne }j\end{array}}(t_j-t_k)}=\frac{1}{{\prod }_{k\mathrm{\ne }j}(t_j-t_k)}\mathrm{.}$ For $j=0$, only the second recursive term contributes and its minus sign supplies the missing factor $t_0-t_q$; for $j=q$, only the first term contributes and division supplies $t_q-t_0$. Their coefficients are also $1\mathrm{/}{\prod }_{k\mathrm{\ne }j}(t_j-t_k)$. This is exactly the coefficient in step 2.1.