This question is inspired by the cute answer to Order of general- and special linear groups over finite fields.
The formula $$ |\mathrm{GL}_n(\mathbb F_q)|=q^{\frac{n(n-1)}2}(q-1)(q^2-1)\cdots(q^n-1). $$ is obtained in that answer by simply but cleverly counting the number of linearly independent $n$-tuples of vectors in $\mathbb F_q^n$.
On the other hand, we know a lot about the structure of the group $\mathrm{GL}_n(\mathbb F_q)$: we can choose (in many ways) a maximal torus; let us take the one consisting of all invertible diagonal matrices, which is a subgroup of order $(q-1)^n$. We can next locate the unipotent radical of the corresponding Borel subgroup which in our case is the subgroup of all upper triangular matrices with $1$s along the main diagonal, thus has order $q^{\frac{n(n-1)}2}$. This accounts for $q^{\frac{n(n-1)}2}(q-1)^n$ elements.
How to account for the remaining factors $\frac{q^2-1}{q-1}$, $\frac{q^3-1}{q-1}$, ..., $\frac{q^n-1}{q-1}$? Do they also correspond to some subgroups that can be named, or maybe some explicitly describable conjugacy classes?
Let $F_q$ be a finite field of . $q$.
Let $T_n(F_q)$ the set of upper triangular matrices (the semi-dircet product of digoanl matrices and unipotent radical).
The quotient $GL(n,F_q)/ T_n(F_q)$ it the set of complete flags $0\subset E_1\subset E_2 \subset ..E_{n-1}\subset E_n=K^n$, where $E_1$ is a 1 dimensional subspace, $E_k$ a $k$-dimensional subsapce), so that for every $i$, $E_{i-1}$ is an hyperplane in $E_{i}$.
Then $q^i-1\over q-1$ appears as the number of $i-1$ dimensional subspace contained in a given $i$ dimensional space : indeed the number of hyperplane in a $d$ dimensionnel vector space over $F_q$ has cardinality $q^d-1\over q-1$ The product $1.{q^2-1\over q-1}... {q^{n-1}-1\over q-1}$ is of course the number of flags.