We prove by induction on a that every positive integer can be written as a product of primes.

The case a = 2 is trivial. So suppose that a is at least 3. If a is a prime then we are done. If a is not a prime then it has a divisor b such that 1 < b < a. According to the induction hypothesis b and a/b can both be written as a product of primes: b = p₁ ··· p_r , a/b = p_{r + 1} ··· p_s. For a we have a = p₁ ··· p_r · p_{r + 1} ··· p_s.

Here we also use induction. The case a = 2 is easy. Suppose that a > 2, and also suppose that uniqueness has been proven for the integers < a. If

a = p₁ ··· p_r   and   a = q₁ ··· q_s

are two ways of expressing a as a product of primes, then it follows that

p₁ | p₁ ··· p_r = q₁ ··· q_s.

From the corollary we may conclude that there exists an index k in the set {1, ... ,s} such that p₁ | q_k. But then we have p₁ = q_k because q_k is a prime. Now apply the induction hypothesis to the integer a/p₁ with the two expressions as products of primes

a/p₁ = p₂ ··· p_r   and  a/p₁ = q₁ ··· q_{k - 1} · q_{k + 1} ··· q_s.

These factorization of a/p₁ are the same (up to the order of the factors) and therefore the two factorization of a are the same.