Let $|\mathbb F|=q$ where $\mathbb F$ is a field. Then $|GL_n(\mathbb F)|=(q^n-1)(q^n-q)(q^n-q^2)\ldots(q^n-q^{n-1})$. 
Let $|\mathbb F|=q$ where $\mathbb F$ is a field. Now 
$$|GL_n(\mathbb F)|=(q^n-1)(q^n-q)(q^n-q^2)\ldots(q^n-q^{n-1}).$$ 

How it is so?
My attempt
I just viewed in the matrix such that 
$\left(\begin{array}{ccccc}
a_{11}&a_{12}&\cdots&a_{1n-1}&a_{1n}\\
\vdots&&\vdots&&\vdots\\
a_{n1}&a_{n2}&\cdots&a_{nn-1}&a_{nn}
\end{array}\right)$
For the first element I will have q choices and rest of the elements also have q choices. So there are $q^n$ choices and i should neglect $0$ in all at a time. So I removed one choice from the total so for the first row I will have $q^n-1$ choices. But for the second and third i am having confusion how to see the choices. And I am thinking that this will give the number of distinct basis for $GL_n(\mathbb F)$. Is it correct. Give me a diagrammatical explanation.

Hint: We want to choose $n$ linearly independent vectors $v_1,\ldots,v_n$ in $\mathbb{F}^n$. For $v_1$ we can choose any non-zero vector, so we have $q^n-1$ choices. For $v_2$ we can choose any vector which is not a multiple of $v_1$, so we have $q^n-q$ choices (since $q_1 \neq 0$). For $v_3$ we can choose any vector which is not in the span of $v_1,v_2$, and since $v_1,v_2$ are linearly independent, this rules out exactly $q^2$ vectors, leaving $q^n-q^2$ choices. And so on.
