Using permutations explain how for a triangular matrix only one term can be non zero.
Please do not include any proofs using the cofactor method.
Edit (OP's attempt as written in the comment section):
My [attempted] proof states that for a lower triangular matrix we start with $n$ choices but $n-1$ of them are zero. We must choose the upper left hand corner because the other choices will lead to a zero term and therefore a zero permutation. Moving on to the next row we are faced with $n-1$ choices, but $n-2$ of them are zero. Therefore we again must make the choice along the diagonal. The proof continues like this until we arrive at the lower right hand corner. Then I declare that $\det(A)= \det(A')$ and therefore it is proven for any triangular matrix.