Value of Vandermonde Determinant/Formulation 1/Proof 3

Proof
Let:
 * $V_n = \begin {vmatrix}

1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\ 1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1} \end {vmatrix}$

Start by replacing number $x_n$ in $V_n$ with the unknown $x$.

Thus $V_n$ is made into a function of $x$.


 * $\map P x = \begin{vmatrix}

1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\ 1 & x & x^2 & \cdots & x^{n - 2} & x^{n - 1} \end{vmatrix}$.

Let $x$ equal a value from the set $\set {x_1,\ldots,x_{n-1} }$.

Then determinant $\map P x$ has equal rows, giving:

Perform row expansion by the last row.

Then $\map P x$ is seen to be a polynomial of degree $n-1$:


 * $\map P x = \begin {vmatrix}

x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \end{vmatrix} + \begin{vmatrix} 1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ 1 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ 1 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \end {vmatrix} x + \cdots + \begin {vmatrix} 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} \\ 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} \\ 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} \end{vmatrix} x^{n - 1}$

By the Polynomial Factor Theorem:
 * $\map P x = \map C {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$

where $C$ is the leading coefficient (with $x^{n - 1}$ power).

Thus:
 * $\map P x = V_{n - 1} \paren {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$

which by evaluating at $x = x_n$ gives:


 * $V_n = V_{n - 1} \paren {x_n - x_1} \paren {x_n - x_2} \dotsm \paren {x_n - x_{n - 1} }$

Repeating the process:

which establishes the solution.