Let $V_n = \begin{vmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\   1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\   1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$. By Multiple of Row Added to Row of Determinant, we can subtract row 1 from each of the other rows and leave $V_n$ unchanged: :$V_n = \begin{vmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   0 & x_2 - x_1 & x_2^2 - x_1^2 & \cdots & x_2^{n-2} - x_1^{n-2} & x_2^{n-1} - x_1^{n-1} \\   0 & x_3 - x_1 & x_3^2 - x_1^2 & \cdots & x_3^{n-2} - x_1^{n-2} & x_3^{n-1} - x_1^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   0 & x_{n-1} - x_1 & x_{n-1}^2 - x_1^2 & \cdots & x_{n-1}^{n-2} - x_1^{n-2} & x_{n-1}^{n-1} - x_1^{n-1} \\   0 & x_n - x_1 & x_n^2 - x_1^2 & \cdots & x_n^{n-2} - x_1^{n-2} & x_n^{n-1} - x_1^{n-1} \end{vmatrix}$ Similarly without changing the value of $V_n$, we can subtract, in order: :$x_1$ times column $n-1$ from column $n$ :$x_1$ times column $n-2$ from column $n-1$ and so on, till we subtract: : $x_1$ times column $1$ from column $2$. The first row will vanish all apart from the first element $a_{11} = 1$. On all the other rows, we get, with new $i$ and $j$: :$a_{ij} = \left({x_i^{j-1} - x_1^{j-1}}\right) - \left({x_1 x_i^{j-2} - x_1^{j-1}}\right) = \left({x_i - x_1}\right) x_i^{j-2}$: :$V_n = \begin{vmatrix}   1 & 0 & 0 & \cdots & 0 & 0 \\   0 & x_2 - x_1 & \left({x_2 - x_1}\right) x_2 & \cdots & \left({x_2 - x_1}\right) x_2^{n-3} & \left({x_2 - x_1}\right) x_2^{n-2} \\   0 & x_3 - x_1 & \left({x_3 - x_1}\right) x_3 & \cdots & \left({x_3 - x_1}\right) x_3^{n-3} & \left({x_3 - x_1}\right) x_3^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   0 & x_{n-1} - x_1 & \left({x_{n-1} - x_1}\right) x_{n-1} & \cdots & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-3} & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-2}\\   0 & x_n - x_1 & \left({x_n - x_1}\right) x_n & \cdots & \left({x_n - x_1}\right) x_n^{n-3} & \left({x_n - x_1}\right) x_n^{n-2} \end{vmatrix}$ For all rows apart from the first, the $k$th row has the constant factor $\left({x_k - x_1}\right)$. So we can extract all these as factors, and from Determinant with Row Multiplied by Constant, we get: :$\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) \begin{vmatrix}   1 & 0 & 0 & \cdots & 0 & 0 \\   0 & 1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\   0 & 1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   0 & 1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\   0 & 1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2} \end{vmatrix}$ From Determinant with Unit Element in Otherwise Zero Row, we can see that this directly gives us: :$\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) \begin{vmatrix}   1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\   1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\   1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2} \end{vmatrix}$ and it can be seen that: : $\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) V_{n-1}$ $V_2$, by the time we get to it (it will concern elements $x_{n-1}$ and $x_n$), can be calculated directly using the formula for calculating a Determinant of Order 2: : $V_2 = \begin{vmatrix}   1 & x_{n-1} \\   1 & x_n  \end{vmatrix} = x_n - x_{n-1}$ The result follows. {{qed}}