Prime not Divisor implies Coprime/Proof 1

Proof
Let $p \in \Bbb P, p \nmid a$.

We need to show that $\gcd \set {a, p} = 1$.

Let $\gcd \set {a, p} = d$.

As $d \divides p$, we must have $d = 1$ or $d = p$ by GCD with Prime.

But if $d = p$, then $p \divides a$ by definition of greatest common divisor.

So $d \ne p$ and therefore $d = 1$.