Vandermonde Determinant

\[ \begin{aligned} &\det\left(\begin{array}{cccc} x_0^0 & x_1^0 & \cdots & x_{n-1}^0\\ x_0^1 & x_1^1 & \cdots & x_{n-1}^1\\ \vdots & \vdots & \vdots &\vdots\\ x_0^{n-1} & x_1^{n-1} & \cdots & x_{n-1}^{n-1} \end{array}\right) \\ \\ & 第\ i\ 行减去第\ i-1\ 行乘\ x_0\ ，得到 \\ \\ =&\det\left(\begin{array}{cccc} 1 & 1 & 1 & \cdots & 1\\ 0 & x_1^0(x_1-x_0) & x_2^0(x_2-x_0) & \cdots & x_{n-1}^0(x_{n-1}-x_0)\\ 0 & x_1^1(x_1-x_0) & x_2^1(x_2-x_0) & \cdots & x_{n-1}^1(x_{n-1}-x_0)\\ \vdots & \vdots & \vdots &\vdots & \vdots \\ 0 & x_1^{n-2}(x_1-x_0) & x_2^{n-2}(x_1-x_0) & \cdots & x_{n-1}^{n-2}(x_{n-1}-x_0) \end{array}\right) \\ \\ & 于是可以把第一行第一列去掉，然后剩下的每一列提取 x_i-x_0，得到 \\ \\ =&\prod_{i>1} (x_i-x_0) \left(\begin{array}{cccc} x_1^0 & x_2^0 & \cdots & x_{n-2}^0\\ x_1^1 & x_2^1 & \cdots & x_{n-2}^1\\ \vdots & \vdots & \vdots &\vdots\\ x_0^{n-2} & x_1^{n-2} & \cdots & x_{n-2}^{n-2} \end{array}\right) \\ \\ & 简单归纳知道答案就是 \\ =&\prod_i\prod_j (x_i-x_j) \end{aligned} \]