Evaluate $o(GL_2 (\mathbb{Z}/p))$

Question

The order of $GL_2 (\mathbb{Z}/p)$ seems to be $(p^2 - 1)(p^2 - p)$ using the fact stated in this Wiki page.

Quoting a line from there :

This can be shown by counting the possible columns of the matrix : the first column can be anything but the zero vector; the second column can be anything but the multiples of the first column; and in general, the $k$ th column can be any vector not in the linear span of the first $k − 1$ columns

I know that the the number of distinct (unordered) bases of an $n$-$\text{dim}$ vector space over $\mathbb{F}_p$ is $\frac{(q^n - 1)\cdots (q^n - q^{n-1})}{n!}$ (call this Lemma $1$). I have an understanding of its proof as well. I am not very much aware of matrices and I have not worked with them.

$\color{red}{[1]}$ How do I prove this using the Lemma $1$ stated above (and/or using the strategy written on the wiki article)?

$\color{red}{[2]}$ It appears that $\left\{\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} \right\}$ forms a basis of $GL_2(\mathbb{Z/p})$ which makes it $4$-$\text{dim}$. Why do we work with $n=2$ for this then?

Your statement [2] makes no sense at all. ${\rm GL}_2(Z/p)$ is a multiplicative group, not a vector space. — Derek Holt, May 25 '20 at 09:15
@DerekHolt Can we not consider $V = \left{ \begin{pmatrix} a & b \ c & d\end{pmatrix} : a,b,c,d \in \mathbb{(Z/p)}, p \not\mid ad-bd \right}$ over $\mathbb{F}_p$? — oldsailorpopoye, May 25 '20 at 09:58
Yes, we can. But your matrices have determinant zero, so are not in the general linear group. — Dietrich Burde, May 25 '20 at 10:52
@DietrichBurde I wrote $p \not\mid ad - bc$ so as to make $V$ and $GL_n(Z/p)$ isomorphic — oldsailorpopoye, May 25 '20 at 11:24
But you didn't write this in $[2]$. There are still the matrices with $ad-bc=0$. — Dietrich Burde, May 25 '20 at 11:26

Evaluate $o(GL_2 (\mathbb{Z}/p))$

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