Let where . Then can be written uniquely (up to permutation) as a product of prime numbers.
Proof:
We use induction. We know the theorem is true for . Now suppose the theorem is true for every . If is prime, the theorem is also true for . Otherwise, by definition where and . It follows that , so both and can be written as products of prime numbers by our induction hypothesis. Since , we have proven the existence of a prime factorization for every .
Now suppose there are numbers with non-unique prime factorizations. Let be the least such integer. Then for primes and . We know , which by TheoremTheorem means divides some , WLOG, let it be . Since they are both prime, we know . Thus , which contradicts the minimality of .