Product of primitive roots of unity (D&F 13.6.1)

Suppose \(m\) and \(n\) are relatively prime positive integers. Let \(\zeta_m\) and \(\zeta_n\) be primitive \(m\)th and \(n\)th roots of unity. Prove that \(\zeta_m \zeta_n\) is a primitive \(mn\)th root of unity.

This is essentially proving that if \(a\) and \(b\) are integers such that \((a,m) = (b,n) = 1\) then \((an+bm,mn) = 1\) (if the conclusion holds then \((an+bm-kmn,mn) = 1\) for any integer \(k\)). Note that any prime divisor \(d\) of \(an+bm\) cannot appear in the prime factorization of \(m\) or \(n\). For instance if \(d \mid m\) then it follows that \(d \mid an\), and since \((m,n) = 1\) we have \(d \mid a\), a contradiction. Thus the prime factorization of \(an+bm\) is disjoint from that of \(mn\), and the conclusion follows.