Why is the determinant of any triangular matrix always the multiple of the main diagonal?

Question

Is there a mathematical proof or a conclusion explaining as to why it is that?

Answer 1

Expand along the row/column with maximum number of zeroes. You might need induction to prove this.

Answer 2

For an $n\times n$ upper triangular matrix, $$ U=\begin{bmatrix}u_{11} & u_{12}^* \\ 0 & U_{22}\end{bmatrix}, $$ where $U_{22}$ is $(n-1)\times(n-1)$, we have using the Leibniz formula that $$ \det(U)=u_{11}\det(U_{22}). $$ Now use induction.