$n\times n$ matrix with zero determinant?

$n\times n$ matrix with zero determinant?

Let $n\ge 3$ be a natural number, and for $1\leq k\leq n-2$ consider the
$n\times n$ matrix
$$A_{n,k}=\begin{pmatrix} 1^{k}& 2^{k}& \cdots& n^{k}\\ (n+1)^k& (n+2)^k&
\cdots& (2n)^k\\ \vdots& \vdots& & \vdots\\ (n^2-n+1)^k& (n^2-n+2)^k&
\cdots& n^{2k} \end{pmatrix}_{\large{.}}$$
It seems that $\det (A_{n,k})=0$. I can prove this for $n=3$ and $k=1$,
and for all $n\ge 4$ and $k=1,2.$ But my ways do not generalize to higher
values of $k$.
Is there a simple proof that $\textbf{$\det(A_{n,k})=0$?}$