Here $p$ is prime and $GL_d(\mathbb{F})$ denotes the group of invertible $d\times d$ matrices over the field $\mathbb{F}$. I have seen several posts computing the order of $GL_2(\mathbb{Z}/p\mathbb{Z})$ as follows: We know a matrix is invertible iff it has linearly independent columns. So there are $p^2-1$ choices for the first column (all vectors are fair game except the zero vector) and for the second vector, there are $p^2-1-(p-1)$ choices (since any non-zero vector is OK except for non-zero multiples of column one).
I understand that one could continue in this method if $d>2$, but it would get somewhat messy. Surely there must be a simpler method to compute this. Does anyone have any ideas?
This is a straight forward generalization of the $2×2$ case. Same reasoning.
We get $(p^d-1)(p^d-p)\cdots(p^d-p^{d-1})$.