This post shows a way to quickly show that the determinant is multiplicative without getting your hands dirty.
Assuming that it is known that there is, up to scale, only one alternating -linear form (i.e. that ), one can proceed as follows. Given , consider the map , . This is clearly -linear and alternating, whence there exists some such that . Evaluating at the identity matrix gives . Evaluating at gives .
Of course, using the trick similar to the first lemma here, it suffices to show this for to obtain it for any unitary commutative ring, after showing that the determinant is in fact a polynomial equation with integer coefficients (for example, by showing the Leibniz formula).