The Van der Monde determinant associated to numbers \(x_1,\dots ,x_n\), \(n\geq 2\), is
\[ V_n \DefEq \det\, \left[ \begin{array}{ccccc} 1 & x_1 & x_{1}^{2} & \cdots & x_{1}^{n-1} \\ 1 & x_2 & x_{2}^{2} & \cdots & x_{2}^{n-1} \\ \vdots & \vdots & & \ddots & \vdots \\ 1 & x_n & x_{n}^{2} & \cdots & x_{n}^{n-1} \\ \end{array} \right] \]We will use the multilinearity properties of the determinant operation to show
\[ V_n\ =\ \prod_{1\leq j< i \leq n} (x_i-x_j) \]We compute by induction. If \(n=2\), we find
\[ \det\, \left[ \begin{array}{cc} 1 & x_1 \\ 1 & x_2 \end{array} \right] = x_2 - x_1 = \prod_{1\leq j < i \leq 2} (x_i - x_j) \]as claimed. Now let \( n>2 \), and suppose the claim is true for \(n-1\); i.e.
\[ V_{n-1} = \det\, \left[ \begin{array}{ccccc} 1 & x_1 & x_{1}^{2} & \cdots & x_{1}^{n-1} \\ 1 & x_2 & x_{2}^{2} & \cdots & x_{2}^{n-1} \\ \vdots & \vdots & & \ddots & \vdots \\ 1 & x_{n-1} & x_{n-1}^{2} & \cdots & x_{n-1}^{n-1} \\ \end{array} \right] = \prod_{1\leq j < i \leq n-1} \]We need to infer the validity of the stated formula for \(V_n\). Do do this, note
\[ V_{n} = \det\, \left[ \begin{array}{ccccc} 1 & x_1-x_n & x_{1}^{2}-x_nx_1 & \cdots & x_{1}^{n-1}-x_nx_{1}^{n-2} \\ 1 & x_2-x_n & x_{2}^{2}-x_nx_2 & \cdots & x_{2}^{n-1}-x_nx_{2}^{n-2} \\ \vdots & \vdots & & \ddots & \vdots \\ 1 & x_{n-1}-x_n & x_{n-1}^{2}-x_nx_{n-1} & \cdots & x_{n-1}^{n-1}-x_nx_{n-1}^{n-2} \\ 1 & 0 & 0 & \cdots & 0 \end{array} \right] \]obtained by subtracting \(x_n\) times (second last column) from (last column). Then subtract \(x_n\) times (third last column) from (second last column) etc. Expand now along the last row to get
\[ V_{n} = (-1)^{n+1}\det\, \left[ \begin{array}{cccc} x_1-x_n & x_{1}^{2}-x_nx_1 & \cdots & x_{1}^{n-1}-x_nx_{1}^{n-2} \\ x_2-x_n & x_{2}^{2}-x_nx_2 & \cdots & x_{2}^{n-1}-x_nx_{2}^{n-2} \\ \vdots & & \ddots & \vdots \\ x_{n-1}-x_n & x_{n-1}^{2}-x_nx_{n-1} & \cdots & x_{n-1}^{n-1}-x_nx_{n-1}^{n-2} \end{array} \right] \]The first row is a multiple of \((x_1-x_n)\). The second row is a multiple of \((x_2-x_n)\) etc. Thus
| \(V_n\) | \(=\) | \( (-1)^{n+1}(x_1-x_n)\dots (x_{n-1}-x_n)\det \left[ \begin{array}{cccc} 1 & x_1 & \cdots & x_{1}^{n-2} \\ 1 & x_2 & \cdots & x_{2}^{n-2} \\ \vdots & & \ddots & \vdots \\ 1 & x_{n-1} & \cdots & x_{n-1}^{n-2} \end{array} \right] \) |
| \(\) | \(=\) | \( (-1)^{n-1}\left( \prod_{1\leq i < n} (x_i-x_n) \right) V_{n-1} \) |
| \(\) | \(=\) | \( (-1)^{n-1}\left( \prod_{1\leq i < n} (-1)(x_n-x_i) \right) V_{n-1} \) |
| \(\) | \(=\) | \( (-1)^{2(n-1)}\left( \prod_{1\leq i < n} (x_n-x_i) \right) \left( \prod_{1\leq j < i \leq n-1} (x_i-x_j) \right) \) |
| \(\) | \(=\) | \( \prod_{1\leq j < i \leq n} (x_i - x_j) \) |
which is exactly what we wanted to show.