What is the number of invertible $n\times n$ matrices in $\operatorname{GL}_n(F)$? $F$ is a finite field of order $q$. What is the size of $\operatorname{GL}_n(F)$ ?
I am reading Dummit and Foote "Abstract Algebra".  The following formula is given:  $(q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$.  The case for $n = 1$ is trivial.  I understand that for $n = 2$  the first row of the matrix can be any ordered pair of field elements except for $0,0$. and the second row can be any ordered pair of field elements that is not a multiple of the first row.  So for $n = 2$  there are $(q^n - 1)(q^n - q)$ invertible matrices.  For $n\geq 3$, I cannot seem to understand why  the formula works.  I have looked at Sloane's OEIS A002884.  I have also constructed and stared at a list of all $168$ $3\times 3$ invertible matrices over $GF(2)$.  I would most appreciate a concrete and detailed explanation of how say $(2^3 - 1)(2^3 - 2)(2^3 - 2^2)$ counts these $168$ matrices.      
 A: In order for an $n \times n$ matrix to be invertible, we need the rows to be linearly independent.  As you note, we have $q^n - 1$ choices for the first row; now, there are $q$ vectors in the span of the first row, so we have $q^n - q$ choices for the second row.  Now, let $v_1, v_2$ be the first two rows.  Then the set of vectors in the span of $v_1, v_2$ is of the form $\{c_1 v_1 + c_2 v_2 | c_1,c_2 \in F\}$.  This set is of size $q^2$, as we have $q$ choices for $c_1$ and $q$ choices for $c_2$.  Thus, we have $q^n - q^2$ choices for the third row.  Continuing this gives the desired formula.
A: For $n=3$, the third row must not be in the subspace generated by the first two rows.  A vector in this subspace requires $2$ coefficients ($q^2$ possibilities), you  must substract $q^2$ vectors, whence a third factor $q^3-q^2$. And so on.
A: A natural generalization of this question is "What fraction of all matrices in $\mathcal{M}_N(\mathbb{F}_q)$ are invertible?"
The answer can be found by a straightforward generalization of this answer:
the fraction is
\begin{align*}
\frac{|GL_n(\mathbb{F}_q)|}{|\mathcal{M}_N(\mathbb{F}_q)|} &= \frac{\prod \limits_{k=0}^{n-1} (q^n - q^k)}{q^{n^2}} \\
&= \prod_{k=1}^n \left( 1-q^{-k} \right) \\
&= \left(\frac{1}{q}, \frac{1}{q} \right)_n,
\end{align*}
where the last expression denotes the q-Pockhammer symbol.
In the limit $n \to \infty$, this fraction approaches $(1/q, 1/q)_\infty$, which equals $\phi(1/q)$, where $\phi$ is the Euler function.
This limiting fraction is finite for any $q$, but increases monotonically with $q$ and approaches $1$ as $q$ becomes large.
