Theorem: Value of Vandermonde Determinant/Formulation 1

Let $V_n$ be the '''Vandermonde determinant of order $n$''' defined as the following formulation:
{{:Definition:Vandermonde Determinant/Formulation 1}}
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$

Proof:
Let:
:<nowiki>$V_n = \begin {vmatrix}
  1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
  1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
  1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
  1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
  1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1}
\end {vmatrix}$</nowiki>
Start by replacing number $x_n$ in $V_n$ with the unknown $x$.
Thus $V_n$ is made into a function of $x$:
:<nowiki>$\map P x = \begin{vmatrix}
  1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
  1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
  1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
  1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
  1 & x & x^2 & \cdots & x^{n - 2} & x^{n - 1}
\end{vmatrix}$</nowiki>
Let $x$ equal a value from the set $\set {x_1, \ldots, x_{n-1} }$.
Then determinant $\map P x$ has equal rows, giving:
{{begin-eqn}}
{{eqn | l = \map P x
      | r = 0
      | c = for $x = x_1, \ldots, x_{n - 1}$ 
}}
{{end-eqn}}
Perform row expansion by the last row.
Then $\map P x$ is seen to be a polynomial of degree $n-1$:
:<nowiki>$\map P x = \begin {vmatrix}
  x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
  x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
  x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
  x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}
\end{vmatrix} + 
\begin{vmatrix}
  1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
  1 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
  1 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}
\end {vmatrix} x + \cdots + 
\begin {vmatrix}
  1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} \\
  1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} \\
  1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2}
\end{vmatrix} x^{n - 1}$</nowiki>
<!---
We can see that statement $\map P x = 0$ holds true for all $x_1, x_2, \cdots x_{n - 1}$ because if $x = x_1, x_2, \cdots x_{n - 1}$ starting determinant would have two equal rows and by Square Matrix with Duplicate Rows has Zero Determinant would $V_n = 0$.
{{explain|The meaning of the above statement needs to be made clearer, and the logical sense turned into a linear flow.}}
Moved statement higher up so it reads linearly. Nov 6 2019 GBGustafson
Also fixed one typo $x_{n - 1} \to x^{n - 1}$ and changed "returning" to "evaluating at".
The cofactor expansion is missing checkerboard signs but the minors are correct. Not fixed.
--->
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:
{{begin-eqn}}
{{eqn | l = V_n
      | r = \prod_{1 \mathop \le i \mathop < n} \paren {x_n - x_i} V_{n - 1}
      | c = 
}}
{{eqn | r = \prod_{1 \mathop \le i \mathop < n} \paren {x_n - x_i} \prod_{1 \mathop \le i \mathop < n - 1} \paren {x_{n - 1} - x_i} V_{n - 2}
      | c = 
}}
{{eqn | r = \dotsm
      | c = 
}}
{{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}
      | c = 
}}
{{end-eqn}}
which establishes the solution.
{{qed}}