Maximal order with primitive determinant in $\operatorname{GL}_n(\mathbb{F}_q)$

For $A\in GL_n(\Bbb{F}_q)$ if $order(A)=q^n-1$ then $A$'s minimal polynomial $P(x)\in \Bbb{F}_q[x]$ divides $x^{q^n-1}-1$ which is separable so that $A$ is diagonalizable (over the algebraic closure).

Expressing the order of $A$ in term of the lcm of the order of the eigenvalues, that is the roots of $P(x)$, you'll get that $order(A)=q^n-1$ iff the roots of $P$ have order $q^n-1$, which implies that $P$ is irreducible of degree $n$ and that the roots generate $\Bbb{F}_{q^n}^\times$.

Letting $a$ be a root then the others are $a^{q^m},m\in 1\ldots n-1$ (Galois theory, Frobenius automorphism).

Since $\deg(P)=n$ then $P$ is also the characteristic polynomial of $A$, therefore

$$\det(A)= \prod_{m=0}^{n-1} a^{q^m} = a^{(q^n-1)/(q-1)}$$

It has order $q-1$ and it generates $\Bbb{F}_q^\times$.

Conversely take $a\in \Bbb{F}_{q^n}^\times$ of order $q^n-1$, let $P(x)\in \Bbb{F}_q[x]$ be its $\Bbb{F}_q$-minimal polynomial and $A$ its companion matrix. Then $P$ is the characteristic and minimal polynomial of $A$ so that $a$ is an eigenvalue of $A$ which will have order $q^n-1$.