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

Theorem
Let $n$ be an integer such that $n > 1$.

Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.

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 $p_1 \mathop \backslash m \implies p_1 \mathop \backslash q_1 q_2 \ldots q_s$.

Thus $\exists j: 1 \le j \le s: p_1 \mathop \backslash q_j$ 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.