Editing Vandermonde matrix (section)

===First proof: Polynomial properties===

The first proof relies on properties of polynomials. 

By the [[Leibniz formula (determinant)|Leibniz formula]], <math>\det(V)</math> is a polynomial in the <math>x_i</math>, with [[integer]] coefficients. All entries of the <math>(i-1)</math>-th column have [[total degree]] <math>i</math>. Thus, again by the Leibniz formula, all terms of the determinant have total degree 
:<math>0 + 1 + 2 + \cdots + n = \frac{n(n+1)}2;</math>
(that is, the determinant is a [[homogeneous polynomial]] of this degree).

If, for <math>i \neq j</math>, one substitutes <math>x_i</math> for <math>x_j</math>, one gets a matrix with two equal rows, which has thus a zero determinant. Thus, considering the determinant as [[univariate]] in <math>x_i,</math> the [[factor theorem]] implies that <math>x_j-x_i</math> is a divisor of <math>\det(V).</math> It thus follows that for all <math>i</math> and <math>j</math>, <math>x_j-x_i</math> is a divisor of <math>\det(V).</math>

This will now be strengthened to show that the product of all those divisors of <math>\det(V)</math> is a divisor of <math>\det(V).</math> Indeed, let <math>p</math> be a polynomial with <math>x_i-x_j</math> as a factor, then <math>p=(x_i-x_j)\,q,</math> for some polynomial <math>q.</math> If <math>x_k-x_l</math> is another factor of <math>p,</math>  then <math>p</math> becomes zero after the substitution of <math>x_k</math> for <math>x_l.</math> If <math>\{x_i,x_j\}\neq \{x_k, x_l\}, </math> the factor <math>q</math> becomes zero after this substitution, since the factor <math>x_i-x_j</math> remains nonzero. So, by the factor theorem, <math>x_k-x_l</math> divides <math>q,</math> and <math>(x_i-x_j)\,(x_k-x_l)</math> divides <math>p.</math> 

Iterating this process by starting from <math>\det(V),</math> one gets that <math>\det(V)</math> is divisible by the product of all <math>x_i-x_j</math> with <math>i<j;</math> that is 
:<math>\det(V)=Q\prod_{0\le i<j\le n} (x_j-x_i),</math>
where <math>Q</math> is a polynomial. As the product of all <math>x_j-x_i</math> and <math>\det(V)</math> have the same degree <math>n(n + 1)/2</math>, the polynomial <math>Q</math> is, in fact, a constant. This constant is one, because the product of the diagonal entries of <math>V</math> is <math>x_1 x_2^2\cdots x_n^n</math>, which is also the [[monomial]] that is obtained by taking the first term of all factors in <math>\textstyle \prod_{0\le i<j\le n} (x_j-x_i).</math> This proves that <math>Q=1,</math> and finishes the proof.
:<math>\det(V)=\prod_{0\le i<j\le n} (x_j-x_i).</math>