Induct on \(r\). It’s obvious if \(r = 1\), and it is just the statement of Lemma 1.9 when \(r = 2\). So assume it holds for \(r\), and let \(a_1,\ldots , a_{r+1}\) be a nonempty list of integers such that \(p|\prod _{i=1}^{r+1} a_i\). Then, since we know the result when \(r=2\), either \(p | \prod _{i=1}^r a_i\) or \(p | a_{r+1}\). In the first case, we’re done by the induction hypothesis. In the second case, it’s just obvious that we’re done.

As in the first proof, assume, to get a contradiction, that Theorem 1.8 does not hold. Then we can find a positive integer \(n\) with two distinct prime factorizations \(p = [p_1,\ldots , p_r]\) and \(q = [q_1,\ldots , q_s]\). As we noted in the beginning of the first proof, it is easy to see that \([]\) is the unique prime factorization of \(1\). So \(n {\gt} 1\). Moreover, using Axiom 1.3 we can assume that \(n\) is the smallest positive integer with two distinct prime factorizations, and, by switching \(p\) and \(q\) if necessary, we can assume that \(p_1 \leq q_1\). Then \(p_1 | n\). So, by Corollary 1.10, \(p_1 | q_i\) for some \(i\). Since \(q_i\) is prime and \(p_1 {\gt} 1\), this implies that \(p_1 = q_i\). So, then \(p_1 \leq q_1 \leq q_i = p_1\), and this implies that \(p_1 = q_1\). But then \(p_2\cdots p_r = q_1\cdots q_s\). So, by the minimality of \(n\), \(r = s\) and \(p_i = q_i\) for all \(i = 2,\ldots , r\). So, in fact, \(p = q\). This contradicts our assumption that \(n\) has two distinct prime factorizations. So the theorem is proved.