Let $p$ be a prime number and $r$ even such that $0<r<n<p$. Calculate rank of matrix$\begin{pmatrix}1&1&1&...&1\\ 0&1&2&...&n-1\\ 0^2&1^2&2^2&...&(n-1)^2\\ \vdots&\vdots&\vdots&&\vdots\\ 0^{r-1}&1^{r-1}&2^{r-1}&...&(n-1)^{r-1} \end{pmatrix} \in \mathbb Z_p^{r\times n}$
Here is an example for $0<2<3<7$ we have $\begin{pmatrix}1&1&1\\ 0&1&2\\ \end{pmatrix}$. It's rank is $2=r$.
Another $0<4<5<7 $ :$\begin{pmatrix}1&1&1&1&1\\ 0&1&2&3&4\\ 0^2&1^2&2^2&3^2&4^2\\ 0^{3}&1^{3}&2^{3}&3^3&4^{3} \end{pmatrix}=\begin{pmatrix}1&1&1&1&1\\ 0&1&2&3&4\\ 0&1&4&2&2\\ 0&1&1&6&1 \end{pmatrix}\rightarrow \begin{pmatrix}1&0&0&0&6\\ 0&1&0&0&4\\ 0&0&1&0&1\\ 0&0&0&1&4 \end{pmatrix}$ has full rank $4=r$.
Does it hold in general? If so, how do I proove it?
As already observed in the comments, this matrix is related to a Vandermonde matrix. Specifically, the $(1,1)$ minor is the determinant of a matrix that is obtained from a Vandermonde matrix with distinct factors by multiplying each column with a non-zero factor. Thus, the Laplace expansion for the first column yields a non-zero determinant; hence the matrix has full rank.