$\GL(n, \Fq)$ is the group of invertible $n \times n$ matrices over the finite field $\Fq$ Here is a proof that $ \lvert \GL(n, \Fq) \rvert = (q^n - 1)(q^n - q) \dotsm (q^n - q^{n-1}) $

Each element $A \in \GL(n, \Fq)$ is given by a collection of $n$ $\Fq$ linearly independent vectors. If one chooses the first column vector of $A$ from $(\Fq)^n$ there are $q^n$ choices, but one can't choose the zero vector since this would make the determinant of $A$ zero. So there are really only $q^n-1$ choices. To choose an $i$ th vector from $(\Fq)^n$ which is linearly independent from $i-1$ already chosen linearly independent vectors $\lbrace V_1, \dots, V_{i-1}\rbrace$ one must choose a vector not in the span of $\lbrace V_1, \dots, V_{i-1} \rbrace$ There are $q^{i-1}$ vectors in this span, so the number of choices is $q^n - q^{i-1}$ Thus the number of linearly independent collections of $n$ vectors in $\Fq$ is $(q^n - 1)(q^n - q) \dotsm (q^n - q^{n-1})$