Talk:Value of Vandermonde Determinant/Formulation 1/Proof 3

Original text:

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 $V_n = 0$.


 * Moved statement higher up so it reads linearly.
 * 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.--Gbgustafson (talk) 11:03, 6 November 2019 (EST)