While proving some facts about matrix group operations on finite fields, I stumbled across the following question:
What is the order of the group of invertible $n\times n$ matrices over a finite field of prime order $p$?
The answer seems to be $$\left|\mathrm{GL}_n(\mathbb F_p)\right|=\prod_{k=1}^n\left(p^n-p^{k-1}\right)\text,$$ but I have not yet seen a satisfying proof: I took a look at this one, but found it to be somewhat sloppy. Is there a nice formal proof of this fact?
To count invertible $n\times n$ matrices is the same as counting bases (the columns of such a matrix form a basis; in other words: A matrix is invertible if and only if the image of the standard basis is a basis) or in fact just to count linearly independent families of $n$ vectors. Given $k\ge0$ linearly independent vectors $v_1,\ldots,v_k\in V=\mathbb F_q^n$, among the $q^n$ elements of $V$ there are $q^k$ that are in the space spanned by the first $k$ vectors, hence there are only $q^n-q^k$ possible choices for $v_{k+1}$ to produce a larger familiy of linearly independent vectors. So if $a_k$ is the number of linearly independent families of $k$ vectors in $\mathbb F_q^n$, we have $$ a_0=1,\qquad a_{k+1}=(q^n-q^k)a_k$$ and conclude $$ a_k=\prod_{i=0}^{k-1}(q^n-q^i)$$ and specifically $$ |\operatorname{GL}_n(\mathbb F_q)|=a_n=\prod_{i=0}^{n-1}(q^n-q^i)=\prod_{k=1}^{n}(q^n-q^{k-1}).$$