According to Wolfram MathWorld,

(https://mathworld.wolfram.com/GeneralizedVandermondeMatrix.html)
Now keep in mind I think this corresponds to $\mathbb{C}$, so it does not necessarily apply to finite fields.
However, if it is somehow possible to factor the determinant into a product as in the regular Vandermonde matrix, if all terms in the product are nonzero, then the product is nonzero by the closure property of the multiplicative group (all nonzero elements) of a finite field.
Edit: I rewrote this answer twice. The first time because I didn't realize the Wolfram article may or may not correspond to $\mathbb{C}$. The second time because I found a formula for the determinant of the generalized Vandermonde matrix from another stack exchange answer. However, that formula apparently corresponded to some different form of the matrix, that they claimed was also called a generalized Vandermonde matrix, so I don't think my second answer applied to the question here.