Problem:Let $ F_q$ be a finite field with $ q$ elements.

$ T_n(F_q) := \{ A = (a_{ij}) \in F^{n \times n}$ | $ a_{ij} = 0$ for $ i < j,$ and $ a_{ij} \neq 0$ $ \forall i \}$ .

Determine the number of elements in $ T_n(F_q)$ .

My solution is as follows:

Starting with the last row going upwards, there are:

$ q-1$ possibilities for the last row;

$ (q-1)q$ possibilities for the row before the last;

.

.

.

$ (q-1)q^{n-1}$ possibilities for the first row.

Therefore, in total there are $ (q-1)^nq^{\sum_{i=1}^{n-1} i} = (q-1)^nq^{\frac{n(n-1)}{2}}$ elements.

Could you, please, check my solution?