TITLE: Proof of Vandermonde Matrix Inverse Formula
QUESTION [11 upvotes]: I'm working through Exercise 40 from section 1.2.3 of Knuth's The Art of Computer Programming volume 1, but am finding myself unable to produce a rigorous proof, and the one here is suspect and not quite clear enough (for me, at least) in some of the steps; in particular, how to "[identify] the $k$th order coefficient in [the] two polynomials."
The problem is this: given a Vandermonde matrix $[x_j^i]_n$, show that the inverse is given by
$$
[b_{ij}]_n = \left[
\frac{
   \sum_{\substack{1 \leq k_1 < \dotsc < k_{n-j} \leq n\\k_1,\dotsc,k_{n-j} \neq i}} (-1)^{j-1} x_{k_1} \dotsc x_{k_{n-j}}
}{
   x_i \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_i)
}
\right]_n\text{.}
$$
The author gives a hint by stating that the sum in the above numerator is just the coefficient of $x^{j-1}$ in the polynomial $(x_1-x)\dotsc(x_n-x)/(x_i-x)$; and he gives an intermediate result showing the explicit multiplication of the matrix and its inverse as
$$
   \sum_{1 \leq t \leq n}b_{it}x_j^t
   =
   \frac{
      x_j \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_j)
   }{
      x_i \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_i)
   }
   =
   \delta_{ij}\text{.}
$$
The only other hint as to the type of solution he was expecting is a reference to A. de Moivre's The Doctrine of Chances, 2nd edition, pp. 197-199, which deals with polynomial recurrence relations and difference products (available here).
At the least, I was just hoping someone could either verify the proof is correct at proofwiki and possibly fill in exactly how one identifies the $k$th order coefficient in the proof; or perhaps explain a proof strategy as to what steps to take where the intermediate result is obtained at some point before the final result.
Thanks so much for any help.

REPLY [5 votes]: I've just done some editing on the proof of Vandermonde's matrix inverse on proofwiki:
https://proofwiki.org/wiki/Inverse_of_Vandermonde_Matrix
I hope that was what you were asking for :)