Determinant of a triangular matrix

Expansion along the first column. Let $T$ be upper triangular with diagonal entries $t_{11},\ldots,t_{nn}$. In the first column, only $t_{11}$ is nonzero, so
$$\det T = t_{11}\det T_{11},$$
where $T_{11}$ is the $(n-1)\times(n-1)$ upper triangular submatrix obtained by removing the first row and first column, with diagonal entries $t_{22},\ldots,t_{nn}$.

Induction. Repeating the expansion, we get $\det T = t_{11}t_{22}\cdots t_{nn}$.

Lower triangular case. The same argument works by expanding along the first row (or by transposing: $\det T = \det T^T$, and $T^T$ is upper triangular).

Conclusion. For any triangular matrix, the determinant equals the product of the diagonal entries.