The question may have been appeared, but I didn't find here.
In matrix group ${\rm GL}_n(\mathbb{F}_q)$, the largest order of any element is equal to $q^n-1$.
It is easy to show existence of element of order $q^n-1$ (by embedding $\mathbb{F}_{q^n}^*$ in this group).
On the other hand, if $A\in {\rm GL}_n(\mathbb{F}_q)$ is of order $k$, then $A^k=I$, so minimal polynomial of $A$ divides $x^k-1$. I was unable to move to prove that $k\le q^n-1$. How to proceed? Any hint?
Let $A$ be your element of $\operatorname{GL}_n(\mathbb{F}_q)$. Consider the subring generated by $A$ inside the ring of all $n \times n$ matrices. By the Cayley-Hamilton theorem, this is a vector space of dimension at most $n$ since $A$ satifies the degree $n$ polynomial identity $\det(xI-A)=0$. Since this is of dimension at most $n$ over the field $\mathbb{F}_q$, this subring consists of at most $q^n$ elements. Removing $0$ from consideration, it consists of at most $q^n-1$ non-zero elements. Hence the order of $A$ as an element of $\operatorname{GL}_n(\mathbb{F}_q)$ is bounded by $q^n-1$.