I am aware that by letting $x=5^{10n}$, one gets $\frac{x^5-1}{x-1}$, and that this is equal to the polynomial $x^4+x^3+x^2+x+1$, which is an integer. My current thinking is to somehow show that if $x-1$ has $k$ factors, $x^5-1$ has more than $k+1$ factors.
(If the opposite is true, then the fraction is prime, as the only way to get a prime number from a fraction of two non-primes is for the fraction to be of the form $\frac{jP}{j}$ for some non-prime $j$ and some prime $P$. )
Looking at the table in this link http://mersennewiki.org/index.php/5_Minus_Tables verifies this to be true by comparing $5^{50}-1$ and $5^{10}-1$, for example, but this does not amount to a proof.
I also note that I have read about cyclotomic/Aureifeuillian factorisation, with particular reference to this problem: Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number, though I do not believe this to be applicable for this problem.
I am also happy to receive an explanation about why this is not solvable, if it happens to be the case.