I know the point of this post is not the actual problem, but I couldn't help myself thinking about it. It turns out that this is actually a more general phenom

I know the point of this post is not the actual problem, but I couldn't help myself thinking about it. It turns out that this is actually a more general phenomenon. Any matrix for which A[i,j] = A[i-1,j] + A[i,j-1] and that has all ones in the first column, i.e. A[0,j] =1, has unit determinant. Pascals triangle/determinant is just a special case when the first column and first row are all ones. Once you postulate this, you can prove it by induction since after one of your matrix "reductions", the (n-1) x (n-1) dimensional sub-determinant is still of the required form.