Here are the statements sorted into the correct order.
Every integer greater than \(1\) can be expressed as a product of primes.
Suppose that the result is not true.
(We’ll show that this leads to a contradiction.)
Then there must be a minimal counterexample.
That is, there is a smallest number that doesn’t have a prime factorisation, say \(n\).
Any prime has a prime factorisation.
So \(n\) is not prime.
So \(n\) is composite, say \(n = ab\) with \(1 < a\), \(b < n\).
Since \(a\) and \(b\) are smaller than \(n\), they have prime factorisations.
(Remember that \(n\) is the smallest number that doesn’t have a prime factorisation.)
But then \(n\) has a prime factorisation: the product of the prime factorisations of \(a\) and \(b\).
So there is not a counterexample—the result is true.
The product is unique up to reordering the factors.
Suppose that some number \(n\) has two prime factorisations.
Say \(n = p_1 \ldots p_k = q_1 \ldots q_l\), where \(p_1\), …, \(p_k\), \(q_1\), …, \(q_l\) are primes, not necessarily distinct.
Then \(p_1\) divides \(q_1 \ldots q_l\) and \(p_1\) is prime, so \(p_1\) divides \(q_1\) or \(p_1\) divides \(q_2\) or … or \(p_1\) divides \(q_l\).
Since \(p_1\), \(q_1\), …, \(q_l\) are prime, we must have \(p_1 = q_r\) for some \(r\).
Then we can cancel these primes from the products.
We can continue in this way.
So \(p_1\), \(p_2\), …, \(p_k\) are in fact \(q_1\), \(q_2\), …, \(q_l\) in some order.