Proof Verification: Euclid's Lemma

Question

Question: Euclid's Lemma: If $p$ is prime, then whenever $p$ divides a product $ab$, $p$ divides at least one of $a, b$.
Prove Euclid's Lemma and the converse, that any natural number having this property (for any pair $a, b$) must be prime.

Proof of Euclid's Lemma: Here, $p$ is a prime and $p|ab$.

Let's assume $a$ and $b$'s prime factorizations are:
$a = m_1m_2 \dots m_r$
$b = n_1n_1 \dots n_s$

Thus, $ab = m_1m_2 \dots m_rn_1n_1 \dots n_s$ Since $p|ab$ and it's a prime, it must be one of $m_1m_2 \dots m_r$ or $n_1n_1 \dots n_s$

Therefore, $p$ divides at least one of $a, b.$

Converse of Euclid's Lemma: For all $a, b$ if $p$ divides a product $ab$ and it divides at least one of $a,b$, then $p$ is a prime.

Proof: Let, $p$ is not a prime. Let, we can write $p$ as:
$p=m_1n_1$ We know, $p|ab$ and $p$ divides at least one of $a,b$
Let's assume $a$ and $b$'s prime factorizations are:
$a = m_1m_2 \dots m_r$
$b = n_1n_1 \dots n_s$

Thus, $ab = (m_1n_1)m_2 \dots m_rn_2 \dots n_s$
We can see in this case $p|ab$ but separately $p \nmid a$ and $p \nmid b$
Which contradicts our initial statement.

Therefore, p must be a prime.

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