Expression for Integer as Product of Primes is Unique/Proof 2

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
Now, suppose $n$ has two prime factorizations, namely $n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$, where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.

Since $p_1 \backslash q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 = q_j,$ for some $1 \le j \le s$.

This means $p_1 \ge q_1$.

Similarly, since $q_1 \backslash p_1 p_2 \dots p_r$, from Euclid's Lemma for Prime Divisors we have $q_1 \ge p_1$.

Thus, $p_1 = q_1$, so we may cancel these common factors, which gives us $p_2 p_3 \cdots p_r = q_2 q_3 \dots q_s$.

We may repeat this process to show that $p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r$.

If $r < s$, we arrive at $1 = q_{r+1} q_{r+2} \cdots q_s$ after canceling all common factors.

Of course, this is impossible, which means $r = s$.

Thus, $p_1 = q_1, p_2 = q_2, \ldots, p_r = q_s$, which means the two factorizations are identical.

Therefore, the prime factorization of $n$ is unique.