Order of General Linear Group over Galois Field

Theorem
If $F$ is a field with $p$ elements, the order of the general linear group $\text{GL}_n(F)$ is:


 * $\displaystyle \prod_{j \mathop = 1}^n \left({ p^n -p^{j-1} }\right)$

Proof
Let $A=[a_{ij}]_{n,n}$ be a matrix such that $|A|\neq 0$ and $a_{ij}\in F$, where $F$ is a field of finitely many elements: $|F|=p$.

How many such matrices can be constructed?

In order to avoid a zero determinant, the top row of the matrix, $\left\{{a_{1j}}\right\}_{j \mathop = 1,\dots,n}$ must have at least one non-zero element.

Therefore, there are $p^n-1$ possibilities for the top row: the $p^n$ possible sequences of $n$ values from $F$, minus the one sequence $0,0, \dots, 0$.

The only restriction on the second row is that it not be a multiple of the first.

Therefore, there are the $p^n$ possible sequences again, minus the $p$ sequences which are multiples of the first row.

Continuing in this fashion, then, the $j^{th}$ row could be any of the $p^n$ possible sequences, minus the $p^{(j-1)}$ sequences which are linear combinations of previous rows.

The number of possible matrices satisfying the conditions of $A$, then, is:


 * $\displaystyle \prod_{j \mathop = 1}^n \left({ p^n -p^{j-1} }\right)$