computation of the order of $\mathrm{GL}(n,{𝔽}_q)$

linearly independent

$\mathrm{GL}(n,{𝔽}_q)$ is the group of invertible $n\times n$ matricesover the finite field ${𝔽}_q$.Here is a proof that$|\mathrm{GL}(n,{𝔽}_q)|=(q^n-1)(q^n-q)\mathrm{\cdots }(q^n-q^{n-1})$.

Each element $A\in \mathrm{GL}(n,{𝔽}_q)$ is given by a collection of $n$${𝔽}_q$-linearly independent vectors (http://planetmath.org/LinearIndependence).If one chooses the first column vector of $A$ from ${({𝔽}_q)}^n$there are $q^n$ choices, but one can’t choose the zero vectorsince this would make the determinant of $A$ zero.So there are really only $q^n-1$ choices.To choose an $i$-th vector from ${({𝔽}_q)}^n$which is linearly independent from $i-1$ already chosenlinearly independent vectors $\{V_1,\mathrm{\dots },V_{i-1}\}$one must choose a vector not inthe span of $\{V_1,\mathrm{\dots },V_{i-1}\}$.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 ${𝔽}_q$is $(q^n-1)(q^n-q)\mathrm{\cdots }(q^n-q^{n-1})$.