Of course, you can always use the fact that if $p$ is a prime which divides a product $ab$ ,then $p | a$ or $p |b$. But, I'm not sure if you have this result, and the question suggests that you don't. If you want to prove it, you can argue along the following lines--if you know Bezout's identity.
Suppose that $p$ divides $ab$ ,but not $a$. Then, it folows that $gcd(a,p)=1$. Hence, we may express $1=np+ma$. Or $b=nbp+mba$. Now, since we know that $p|ab$, we have that $ab=\alpha p$, and so $a=\frac{\alpha p}{b}$. Hence, by substitution, we have that
$b=nbp+m\alpha p=p(nb+\alpha m)$. Hence, b is divisible by p.
For this particular problem, a=b=n.
Note, that if you do not know Bezout's identity, then you will have no idea how I was able to write 1 as a linear combo of p and a. First note that 1 is the gcd of $p$ and $a$. Hence, we can obtain one by iterating the Euclidian algorithm again and again. Then, we just backwards substitute from the last equation upwards to get the gcd as a linear combo of the two numbers involved. Try it with 5 and 17, for instance. This is Bezout's idenitity.