Let $V_n = \begin{vmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\   1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\   1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$. Start by replacing number $x_n$ in $V_n$ with the unknown $x$. Thus $V_n$ is made into a function of $x$. :$P \left({x}\right) = \begin{vmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\   1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\   1 & x & x^2 & \cdots & x^{n-2} & x^{n-1} \end{vmatrix}$. Let $x$ equal a value from the set $\set {x_1,\ldots,x_{n-1} }$. Then determinant $\map P {x}$ has equal rows, giving: {{begin-eqn}} {{eqn | l = \map P {x}       | r = 0       | c = for $x = x_1, \ldots, x_{n-1}$  }} {{end-eqn}} Perform row expansion by the last row. Then $P \left({x}\right)$ is seen to be a polynomial of degree $n-1$: :$P \left({x}\right) = \begin{vmatrix}   x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\   x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\   x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \end{vmatrix} +  \begin{vmatrix}   1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   1 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\   1 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\   1 & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \end{vmatrix}x \ \ + \ \ \cdots \ \ + \ \ \begin{vmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n-2} \\   1 & x_3 & x_3^2 & \cdots & x_3^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\   1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} \end{vmatrix}x^{n-1}$. <!--- We can see that statement $P \left({x}\right) = 0$ 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 $V_n = 0$. {{explain|The meaning of the above statement needs to be made clearer, and the logical sense turned into a linear flow.}} Moved statement higher up so it reads linearly. Nov 6 2019 GBGustafson Also fixed one typo $x_{n-1} \to x^{n-1}$ and changed "returning" to "evaluating at". The cofactor expansion is missing checkerboard signs but the minors are correct. Not fixed. ---> By the Polynomial Factor Theorem: :$P \left({x}\right) = C \left({x - x_1}\right) \left({x - x_2}\right) \dotsm \left({x - x_{n-1} }\right)$ where $C$ is the leading coefficient (with $x^{n-1}$ power). Thus: :$P \left({x}\right) = V_{n-1} \left({x - x_1}\right) \left({x - x_2}\right) \dotsm \left({x - x_{n-1} }\right)$ which by evaluating at $x = x_n$ gives:  :$V_n = V_{n-1} \left({x_n - x_1}\right) \left({x_n - x_2}\right) \dotsm \left({x_n - x_{n-1} }\right)$ Repeating the process: {{begin-eqn}} {{eqn | l = V_n       | r = \prod_{1 \mathop \le i \mathop < n} \left({x_n - x_i}\right) V_{n-1}       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop < n} \left({x_n - x_i}\right) \prod_{1 \mathop \le i \mathop < n-1} \left({x_{n-1} - x_i}\right) V_{n-2}       | c =  }} {{eqn | r = \dotsm       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right)       | c =  }} {{end-eqn}} which establishes the solution. {{qed}}