I calculated that the number of rank $k$ matrices in $\mathbb{F}_q^{n \times m}$ (where we may assume $n \leq m$) for a prime power $q$ is equal to
$$W_{n,m}(k) := \prod_{i = 1}^k \frac{(q^m - q^{i-1})(q^n - q^{i-1})}{(q^k - q^{i-1})}$$
And so summing all $W_{n,m}(k)$ from rank $k = 0$ to full rank $k = n$ should gives us all matrices of $\mathbb{F}_q^{n \times m}$
$$\sum_{k = 0}^n W_{n,m}(k) = q^{nm}$$
Can we also show this equality using pure arithmetic and combinatorial tricks, i.e. using just the formulas for $W_{n,m}$?
I managed to solve this, but it took a number of pages. The full answer can be found here.
In short: one can show that the following $q$-analogue of Pascal's rule holds: $$\prod_{i=1}^k \frac{q^{n+1} - q^{i-1}}{q^k - q^{i-1}} = q^k \cdot \prod_{i=1}^k \frac{q^{n} - q^{i-1}}{q^k - q^{i-1}} + \prod_{i=1}^{k-1} \frac{q^{n} - q^{i-1}}{q^{k-1} - q^{i-1}}$$
Then, this allows you to show that $$ W_{n+1,m}(k) = q^k\cdot W_{n,m}(k) + (q^m - q^{k-1})\cdot W_{n,m}(k-1) $$ by multiplying both sides with $\prod (q^m - q^{i-1})$.
Lastly, use induction on $n$ to get the final equality.
(Fun challenge, try to find the linear algebra interpretation of each statement)