Proof of Vandermonde Matrix
Vandermonde matrix：
$$ \begin{vmatrix}1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{vmatrix} = \prod_{n \ge i > j \ge 1}(x_i-x_j) $$
The proof is as follows：
Starting at line $n$, subtract $x_1$ times the previous line from the subsequent line in the determinant to obtain the following equation:
$$ \begin{align*} \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & x_2-x_1 & \cdots & x_n-x_1 \\ 0 & x_2(x_2-x_1) & \cdots & x_n(x_n-x_1) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & x_2^{n-2}(x_2-x_1) & \cdots & x_n^{n-2}(x_n-x_1) \end{vmatrix}\\ \ \\ \ \\ = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & x_2-x_1 & \cdots & x_n-x_1 \\ 0 & x_2(x_2-x_1) & \cdots & x_n(x_n-x_1) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & x_2^{n-2}(x_2-x_1) & \cdots & x_n^{n-2}(x_n-x_1) \end{vmatrix} \end{align*} $$
Expanding the resulting determinant by the $1$th line.
$$ \begin{align*} &\begin{vmatrix} x_2-x_1 & x_3-x_1 & \cdots & x_n-x_1 \\ x_2(x_2-x_1) & x_3(x_3-x_1) & \cdots & x_n(x_n-x_1) \\ \vdots & \vdots & \ddots & \vdots \\ x_2^{n-2}(x_2-x_1) & x_3^{n-2}(x_3-x_1) & \cdots & x_n^{n-2}(x_n-x_1) \end{vmatrix} \\ \ \\ \ \\ &= (x_2-x_1)(x_3-x_1) \cdots (x_n-x_1) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_2 & x_3 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_2^{n-2} & x_3^{n-2} & \cdots & x_n^{n-2} \end{vmatrix} \end{align*} $$
At this point the right end of the equation is a Vandermonde matrix of order $n-1$, which can be written as
$$ (x_3-x_2)(x_4-x_2) \cdots (x_n-x_2) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_3 & x_4 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_3^{n-3} & x_4^{n-3} & \cdots & x_3^{n-3} \end{vmatrix} $$
According to mathematical induction, the Vandermonde matrix can be written:
$$ \begin{align*} &(x_2-x_1)(x_3-x_1) \cdots (x_n-x_1)(x_3-x_2)(x_4-x_2) \cdots (x_n-x_{n-1}) \\ \ \\ \ \\ &= \prod_{n \ge i > j \ge 1}(x_i-x_j) \end{align*} $$
Q.E.D.