Proof of Euclids' lemma

Question

If $p$ is a prime number and $p \mid ab$ then $p \mid a$ or $p\mid b$, for all integers $a$ and $b$.

The proof starts with:

Without loss of generality, suppose $\gcd(p,a)=1$ (otherwise we are done)

Why are we done when $\gcd(p,a) ≠ 1?$

Answer 1

Since $p$ is a prime number, if the $\gcd(p,a)\neq 1$ then it must be the case that $\gcd(p,a)= p$, which means in particular that $p|a$. Since that was one of the conclusions we were shooting for, that case of the proof is handled. Thus, we skip that part by saying "without loss of generality suppose $\gcd(p,a)= 1$".

Answer 2

To prove that $p\mid a$ or $p\mid b$ is to prove that if $p\nmid a$ then $p\mid b$. And, since $p$ is prime, to assert that $p\nmid a$ is to assert that $\gcd(p,a)=1$.