Let $z_{1},\dots,z_{k}$
be distinct complex numbers with $\left|z_{j}\right|=1,\;j=1,\dots,k$. For any natural $N\geqslant k$
consider the rectangular Vandermonde matrix
$$
V_{N}=\begin{pmatrix}1 & 1 & \dots & 1\\
z_{1} & z_{2} & & z_{k}\\
\vdots & \vdots & & \vdots\\
z_{1}^{N-1} & z_{2}^{N-1} & & z_{k}^{N-1}
\end{pmatrix}.
$$
Let $V_{N}^{*}$ denote the conjugate transpose of $V_{N}$. Since
$V_{N}$ has full rank, the square $k\times k$ matrix $V_{N}^{*}V_N$
is nonsingular. We are interested in the quantity
$$
D\left(N\right)=\det\left(V_{N}^{*}V_N\right).
$$
For $N=k$ we have by the well-known explicit formula
$$
D\left(k\right)=\prod_{i<j}\left|z_{i}-z_{j}\right|^{2}.
$$
Question: Does there exist an explicit “simple'' expression
for $D\left(N\right)$ with arbitrary $N>k$?
Example of a simple expression in the special case $k=2$:
Without loss of generality $z_{1}=1$
and $z_{2}=\exp\left(\imath x\right)$ for $x\in\left[-\pi,\pi\right]\setminus\{0\}.$
An explicit computation gives the following:
$$
D\left(N\right)=N^{2}-\frac{\sin^{2}\frac{N}{2}x}{\sin^{2}\frac{x}{2}}.
$$
Best Answer
This is a partial answer only - but there is a clear place to start.
The Cauchy-Binet theorem gives the answer as a sum of products of all maximal minors of the two matrices, where the minors are taken with matching sets of rows/columns. All of these minors are themselves ordinary, N by N Vandermonde determinants. So the answer is the symmetrization of the squared Vandermonde determinant, over all possible choices of N of the k variables.
This is probably not simple enough for you, but at least it's a formula. I haven't assumed anything about the z variables, nor attempted to do any simplification.