Prove $p$ has only $2$ divisors: $1$ and $p$

Question

How do I prove that

$$ m \ | \ p \iff m = 1 \ \text{or} \ m = p $$ if

$p$ is a natural number and for all natural numbers $k$ and $l$ it is the case that:

$$ p \ | \ kl \ \implies \ p \ |\ k \ \text{or} \ p\ | \ l$$ (definition of the prime number)

Answer 1

Suppose that there is in fact a number $1<m<p$ which divides $p$. What would be a good choice of $k$ and $l$ to show that the second property cannot hold?

Answer 2

Suppose $p=ab$, $a, b \neq 1, p$. Then $p | ab$. If $p|a$, then the fact that $a|p$ implies $a = p$, contradiction. The case $p|b$ is similar.