= Solution
Suppose $n=p_1\cdots p_r=q_1\cdots q_s$ are two products of positive <primes>. If $r=0$, then $n=1$, and necessarily $s=0$, since any nonempty product of positive <primes> is at least $2$. Otherwise <Euclid's lemma>, applied repeatedly to the right-hand product, says $p_1$ divides one $q_j$. Since $q_j$ is <prime>, $p_1=q_j$. Reorder the right-hand factors and cancel this common nonzero factor. Repeating, or using <induction> on the number of factors, matches every factor. Neither side can have factors left over, because their product would then equal $1$. Thus \b[the <prime factorization> is unique up to rearrangement].
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