Expression for Integer as Product of Primes is Unique/Proof 1

Proof
Suppose the supposition false, that is, there is at least one number that can be expressed in more than one way as a product of primes.

Let the least of these be $m$.

That is, $m$ is the smallest number which can be expressed in at least two ways, that is:
 * $m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$

where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime.

Clearly $m$ can not be prime, therefore $r, s \ge 2$.

Let us arrange that the primes are in order of size, that is:
 * $p_1 \le p_2 \le \dots \le p_r$

and:
 * $q_1 \le q_2 \le \dots \le q_s$

and also let us arrange that $p_1 \le q_1$.

Now suppose $p_1 = q_1$. Then:


 * $\dfrac m {p_1} = p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s = \dfrac m {q_1}$

But then we have the integer $\dfrac m {p_1}$ being expressible in two different ways, thus contradicting the fact that $m$ is the smallest number that can be so expressed.

Therefore, $p_1 \ne q_1 \implies p_1 < q_1 \implies p_1 < q_2, q_3, \ldots, q_s$ as we arranged them in order.

Now: $1 < p_1 < q_j: 1 < j < s \implies p_1 \nmid q_j$ from Prime not Divisor implies Coprime.

But $p_1 \mathrel \backslash m \implies p_1 \mathrel \backslash q_1 q_2 \ldots q_s$.

Thus $\exists j: 1 \le j \le s: p_1 \mathrel \backslash q_j$ from Euclid's Lemma for Prime Divisors.

But $q_j$ was supposed to be a prime.

This gives us our contradiction, and there is therefore no such counterexample to our supposition.