Determining the leading coefficient of Vandermonde's Determinant.

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On Shilov's Book "Linear Algebra", when calculating the Vandermonde's Determinant, the author concludes the leading coefficient of the product of the roots of the determinant (seen as a polynomial) is

$$ W(x_1, ..., x_{n-1}) $$

Shilov proceded to see the solution of the determinant as a polynomial of degree $(n - 1)$ in $x_n$, noting that from the definition of Vandermonde's determinant, we have vanishing cases when $x_n$ is equal to any other parameter in $W$ (any determinant with two equal columns vanishes). It makes sense to propose the solution as (polynomial remainder theorem)

$$ W(x_1, .., x_{n}) = k\prod_{k = 1}^{n - 1}(x_n - x_k)$$

with

$$ W(x_1, \ldots, x_n) = \left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}\\ \end{array} \right| $$

Then the author states $$ k = W(x_1,...,x_{n−1})$$ can be found when expanding the las column of the Vandermonde's determinant.

My question is, how did he conclude that?

When expanding $W$ with respect to the last column we do have

$$ W = 1 \cdot A_{1n} - x_nA_{2n} + ... + x_n^{n - 1}W(x_1, ..., x_{n - 1})$$

I believe he then compares this two expressions:

$$ k\prod_{k = 1}^{n - 1}(x_n - x_k) = 1 \cdot A_{1n} + x_nA_{2n} + ... + x_n^{n - 1}W(x_1, ..., x_{n - 1})$$

What's the reason he only takes the last co-factor to determine the value of $k$?

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To make it a bit more clear, let's consider $$W(x_1,\ldots, x_{n-1},X) $$

which is indeed a polynomial in indeterminate $X$ of degree $n-1$ with roots $x_1,\ldots, x_{n-1}$.

By polynomial factor theorem, there is a constant $k$ such that

$$ W(x_1,\ldots, x_{n-1},X) =k \prod_{i=1}^{n-1} (X-x_i). $$

We note that $k$ is the leading coefficient of this polynomial.

On the other hand, by performing expansion by the last column in determinant $W(x_1,\ldots, x_{n-1},X)$ (just like you did, but with $X$ instead of $x_n$), we see that the leading coefficient of polynomial $W(x_1,\ldots, x_{n-1},X) $ is exactly $W(x_1,\ldots, x_{n-1})$ (all expansion cofactors are of course functions of $x_1,\ldots, x_{n-1}$ only and they are the coefficients of the polynomial in $X$).

As we are referring to the leading coefficient of the same polynomial, this implies:

$$k= W(x_1,\ldots, x_{n-1}).$$