Finding the order of $\mathrm{GL}_n(\mathbb{F}_p)$ Is there a way to find the order of the group $\mathrm{GL}_n(\mathbb{F}_p)$.
In my notes for $\mathrm{GL}_3(\mathbb{F}_2)$ it is done by brute force but this does seem like a very good method.
 A: The order of $GL(n,p)$ is:$$(p^n - 1)(p^n - p)(p^n - p^2)\ \cdots\ (p^n - p^{n-1}).$$
This can be shown by counting the possible columns of the matrix: the first column can be anything but the zero vector; the second column can be anything but the multiples of the first column; and in general, the $k$-th column can be any vector not in the linear span of the first $k − 1$ columns.
A: This group acts on the $n$-vectors: the latter set having $p^n$ elements, out of which the zero vector is an orbit and all the non-zero vectors form a single orbit (this translates to: given two non-zero vectors there is a non-singular matrix sending one to the other by matrix multiplication).
SO one can use orbit-stabilizer formula: Take the vector to be $e_1$, the vector with 1 at the top and zero at all other components. The stabilizer subgroup  of this vector is easy to describe: all non-singular matrices with $e_1$ as its first column. Expanding the determinant along the first column we see that the $(n-1)\times(n-1)$ matrix obtained by omitting the first column and the first row has to be non-singular, and the entries on the first row after $a_{11}$ can be arbitrary
Therefore by orbit-stabilizer formula: $|GL(n, F_p)|=(p^n-1)\times p^{n-1}\times |GL(n-1,F_p)|$. Now induction on $n$ along with the fact that $GL_1$ is of order $p-1$ gives the order.
A: Hint: There are as many elements in $GL(n,\mathbb{F}_q)$ as the number of bases of the vectorial space $\mathbb{F}_q^n$.
