= Solution
Let $V_m=\det[x_i^{j-1}]_{i,j=1}^m$. It is an <alternating polynomial>: exchanging two variables exchanges two rows of the <Vandermonde matrix> and reverses its <determinant>. Hence $V_m$ vanishes whenever $x_k=x_l$, so each factor $x_l-x_k$ divides it. The distinct linear factors are pairwise <coprime polynomials>, and their product therefore divides $V_m$. Both have total degree $0+1+\cdots+(m-1)=m(m-1)/2$, so
$$
V_m=C_m\prod_{k<l}(x_l-x_k).
$$
The coefficient of $x_m^{m-1}$ in the <determinant> is $V_{m-1}$, by expansion along its last row. The same coefficient in the product is $\prod_{k<l<m}(x_l-x_k)$. Thus $C_m=C_{m-1}$, and $C_1=1$. Consequently the <Vandermonde determinant> identity is
$$
\boxed{\det[x_i^{j-1}]_{i,j=1}^m=\prod_{1\leq k<l\leq m}(x_l-x_k).}
$$
This is a <polynomial> identity, including repeated coordinates; no division by a possibly zero numerical Vandermonde product is needed.
Back to article page
