Prove that $|GL_n(\mathbb{F})|< q^{n^2}$. Let $\Bbb F$ be a finite field, say $|\Bbb F|=q$; then we know that $|GL_n(\Bbb F)| < \infty$. 

But how can we prove that $|GL_n(\mathbb{F})|< q^{n^2}$? 

I'm guessing because there $n^2$ entries in a $n\times n$ matrix. A little background would be helpful for me about the proof as I am inexperience with algebra proofs. 
 A: While the other answers lead you to the precise order of $GL_n(\mathbf{F})$, I'll explain a simple way to see the inequality you asked about.
As you note, any $n \times n$ matrix has $n^2$ entries; if these entries are coming from $\mathbf{F}$, then the set of all $n \times n$ matrices has order $q^{n^2}$, since there are $q$ choices for each of the $n^2$ entries. Thus, we have $|M_n(\mathbf{F})| = q^{n^2}$.
But the subgroup $GL_n(\mathbf{F})$ of invertible matrices is smaller -- not every $n \times n$ matrix is invertible! For instance, the $n \times n$ zero matrix is not invertible. So we see that $GL_n(\mathbf{F})$  is strictly smaller than $M_n(\mathbf{F})$. So we can conclude that 
$$|GL_n(\mathbf{F})| <  q^{n^2}.$$
A: You have to count the number of independent columns whose entries are not all equal to zero. For the first one you have $q^n-1$ choices. Fixed this one, for the second one you have to count all the possible choices but the multiples of the first column: $q^n-q$.
So till now you have $(q^n-1)(q^n-q)$ choices.
For the third one you have to consider all the $q^n$ choices but the possible combination of the first two columns; these possibles are $q^2$ thus for the third column you have $q^n-q^2$ possibilities.
Hence for the first three columns you have $(q^n-1)(q^n-q)(q^n-q^2)$ possibilities.
Going on till the $n$-th column, you easily get that all the possible choices for $n$ independent columns are $(q^n-1)(q^n-q)(q^n-q^2)\dots(q^n-q^{n-1})$, which is exactly the cardinality of $GL_n(\Bbb F)$.
A: In fact, if $F$ is a field (you weren't clear about what $F$ is) such that $|F|=q<\infty$, then $\left|\text{GL}_n(F)\right|=\left(q^n-1\right)\left(q^n-q\right)\left(q^n-q^2\right)\cdots\left(q^n-q^{n-1}\right)$ (which clearly means $\left|\text{GL}_n(F)\right|<q^{n^2}$).  See http://www-math.mit.edu/~dav/genlin.pdf.
