proof of determinant of the Vandermonde matrix

To begin, note that the determinant of the $n\times n$ Vandermondematrix (which we shall denote as ‘$\mathrm{\Delta }$’) is a homogeneouspolynomial of order $n(n-1)/2$ because every term in the determinantis, up to sign, the product of a zeroth power of some variable times the firstpower of some other variable , $\mathrm{\dots }$, the $n-1$-st power of somevariable and $0+1+\mathrm{\cdots }+(n-1)=n(n-1)/2$.

Next, note that if $a_i=a_j$ with $i\ne j$, then $\mathrm{\Delta }=0$because two columns of the matrix would be equal. Since $\mathrm{\Delta }$ is apolynomial, this implies that $a_i-a_j$ is a factor of $\mathrm{\Delta }$.Hence,

where C is some polynomial. However, since both $\mathrm{\Delta }$ and theproduct on the right hand side have the same degree, $C$ must havedegree zero, i.e. $C$ must be a constant. So all that remains is thedetermine the value of this constant.

One way to determine this constant is to look at the coefficient ofthe leading diagonal, ${\prod }_n{(a_n)}^{n-1}$. Since it equals 1 in boththe determinant and the product, we conclude that $C=1$, hence