1.

AB=I

det(AB)=det(I)

det(A)det(B)=det(I)

det(I)=1.

If det(A) v det(B)=0, then you get zero times another scalar, which is zero.

Hence, det(A) and det(B) ≠ 0.

QED

2.

If AB is singular, then,

det(AB)=0.

So, det(A)det(B)=0.

If neither scalars det(A) or det(B)≠0, then neither one of them could be singular, sinceall nonsingular matricies have nonzero determinants; so, det(A)det(B) wouldn't be equal to zero.

So, A is singular, or B is singular.

QED

Recall that for all scalars a b in R ab=0 iff (a=0)v(b=0)