 Solution

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.