n-rows submatrices of a Vandermonde matrix have full rank?

Question

In this question, they mentioned that $m\times n$ Vandermonde matrices have the property that any submatrix with $n$ rows has full rank.

Does that mean if the submatrix is $n\times r$, where $r\leq n$, then the submatrix has full column rank?

Answer 1

The determinant of any such submatrix formed by the rows $k_1,k_2,\dots k_n$ is $$ \prod\limits_{1\le j<i\le n}(x_{k_i}-x_{k_j}) $$ with $x_{k_i}\neq x_{k_j}$ for $i\neq j$, which is obviously non-zero.