$5^n$ is relatively prime with 13, n in $\mathbb{N}$

Question

$5^n$ is prime with 13, n in $\mathbb{N}$?

I have proved that $5^{n+4}-5^n \equiv$ 0(mod 13) . So $5^n(5^4-1) \equiv$ 0(mod 13)

Now Im stuck on how to prove that $5^n$ is relatively prime with 13

Answer 1

You need to show, that $13$ doesn't divide $5^n$ for every $n\in \mathbb{N}$. Let us assume, that $13$ does divide $5^n$. Therefore, $13$ is a factor of $5^n$. But as 13 is prime and $5$ is the only prime-factor of $5^n$, $13$ cannot divide $5^n$ by the uniqueness of the prime decomposition.