Let $a_1,a_2,\ldots,a_k$ be elements of the Galois field of order $q^n$, say $\mathbb{F}_{q^n}$, with $n \geq k$. 

By Lemma 3.51 of Lidl, Niederreiter-Finite Fields,

$\left|
       \begin{array}{cccc}
         a_1 & a_1^{q} & \cdots & a_1^{q^{k}} \\
         a_2& a_2^{q} & \cdots & a_2^{q^{k}} \\
         \cdots & \cdots& \cdots & \cdots \\
         a_k & a_k^{q} & \cdots & a_k^{q^{k}} \\
       \end{array}
     \right|
=0$ 
if and only if $\{a_1,a_2,\ldots,a_k\}$ are linearly dependent over $\mathbb{F}_q$.

My question is:

is it true that:

$\left|
       \begin{array}{cccc}
         a_1 & a_1^{q^{h_1}} & \cdots & a_1^{q^{h_{k-1}}} \\
         a_2 & a_2^{q^{h_1}} & \cdots & a_2^{q^{h_{k-1}}} \\
         \cdots & \cdots& \cdots & \cdots \\
         a_k & a_k^{q^{h_1}} & \cdots & a_k^{q^{h_{k-1}}} \\
       \end{array}
     \right|
=0$ 

if and only if $\{a_1,a_2,\ldots,a_k\}$ are linearly dependent over $\mathbb{F}_{q^{m}}$ where 
$$m=GCD(h_1,h_2,\ldots,h_{k-1},n)?$$

>**Question.** It is obvious that if they are dependent, then that determinant is $0$. But, is the converse true as well?