Does there exist an explicit criterion (or a good sufficient condition) for proving that a Vandemonde matrix:
$$A(x_1,\dots,x_n):=\left[ \begin{array}{llll}1 & x_1 &\dots& x_1^{n-1}\\ 1 & x_2 &\dots& x_2^{n-1}\\&&\vdots&\\ 1& x_n &\dots& x_n^{n-1}\end{array}\right]$$
has nonzero permanent? ($x_1,\dots,x_n$ are complex numbers) I'm especially interested in the case where $x_i$'s are roots of unity.

**Added**. Here is an explicit conjecture I want to prove:
>If $x_1,\dots,x_n$ are in $\mu_N$ (not necessarily different) and $n$ is coprime to $N$ then $A(x_1,\dots,x_n)\neq 0$.

The conjecture is trivially true when $N$ is a prime and $n<N$, since $A(x_1,\dots,x_n)$ is a sum of $n!$, $N$th roots of unity and $N$ dose not divide $n!$.

**Edit.** I found counterexamples for the above conjecture for a prime $N$ and every $n>N$ s.t. $N\nmid (n+1)$. Explicitly $A(x_1,\dots,x_n)=0$ when all of $x_i$'s are 1 except (1 + $n$ mod $N$) of them which are equal to a primitive $N$th root of unity.