## Solution

Here are the statements sorted into the correct order.

#### Existence

Every integer greater than $1$ can be expressed as a product of primes.

Suppose that the result is not true.

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.

#### Uniqueness

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.