Value of Vandermonde Determinant/Formulation 1/Proof 3

Theorem
The Vandermonde determinant of order $n$ is the determinant defined as follows:


 * $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} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$

Its value is given by:
 * $\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right)$

Proof
Let $V_n = \begin{vmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$.

To obtain the value of the Vandermonde's determinant we start by replacing number $x_n$ in the determinant with the unknown $x$ thus making determinant a function of $x$.

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

After performing row expansion by last row we deduce that function $P(x)$ is polynomial of degree $n-1$:

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

We can see that statement $P(x) = 0$ 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 $V_n = 0$. By the Polynomial Factor Theorem we can now write $P(x) = C(x-x_1)(x-x_2)\cdots(x-x_{n-1})$ where C is the leading coefficient (with $x_{n-1}$ power). So we get: $P(x) = V_{n-1}(x-x_1)(x-x_2)\cdots(x-x_{n-1})$ which by returning $x = x_n$ gives:

$P(x) = V_{n-1}(x_n-x_1)(x_n-x_2)\cdots(x_n-x_{n-1})$

Repeating the same process we get:

$\displaystyle V_n = \prod_{1 \leq i<n} (x_n-x_i)V_{n-1} = \prod_{1 \leq i<n} (x_n-x_i) \prod_{1 \leq i<n-1} (x_{n-1}-x_i) V_{n-2} = \cdots = \prod_{1 \le i < j \le n} (x_j-x_i)$

which establishes the solution.