I didn’t really like the cofactor construction of the inverse of a matrix with nonzero determinant, to prove said inverse exists. I’m willing to accept the equation det(MN) = det(M)det(N) on faith, since I am confident I could work that out if I really had to. With that in my pocket, here’s an explanation of the correspondence between nonzero determinant and invertibility.

Suppose M is invertible. Then there is some M^-1 such that MM^-1 = I, the identity matrix. Then 1 = det(I) = det(MM^-1) = det(M)det(M^-1), and 1 cannot be obtained as a product of 0 with anything, so both M and M^-1 have nonzero determinant.

Now suppose M is (square and) noninvertible. Then the kernel of the transformation T that M represents includes some nonzero vector X, and we may build a basis B for our vector space such that B includes X. The matrix M’ for T relative to B includes a column of all 0s, corresponding to the position of X in the ordering of B, so det(M’) = 0. For appropriate change of basis matrices P, P^-1 we have M = PM’P^-1, so det(M) = det(PM’P^-1) = det(P)det(M’)det(P^-1) = 0.