Power Divides power

Observe that $9^n - 1 = (3^n -1 ) ( 3^n + 1 )$ .
We have $\gcd ( 3^n -1 , 3^n + 1 ) = \gcd ( 2 , 3^n -1 ) = 2$ .
Hence, at most one of these terms is divisible by 5.

Then, if $5^n \mid 9^n - 1$ , we must either have $5^n \mid 3^n - 1$ or $5^n \mid 3^n + 1$ .
However, since $5 ^n > 3^n + 1 > 3^n - 1 > 9$ , hence this is not possible. Thus, there are no solutions.

---

This question can be generalized to deal with $p^n \mid a^{2n} -1$ .

$9^n-1=(10-1)^n-1$

if $n=5^k m$ where $m \in \mathbb{Z^+}$ and $m$ is not a multiple of $5$

Case(1) : If $m=2 l+1$ then

$9^n-1=(10-1)^{5^k(2l+1)}-1=10p-1-1=10p-2=2(5p-1)$

It has not a single factor of $5$ .So, no such integer exists.

Case(2) : If $m=2 l$ then

$9^n-1=(10-1)^{5^k 2 l}-1 \\ =10^{5^k 2 l}-{5^k 2 l \choose 1}10^{(5^k 2l-1)}+ \cdots +{5^k 2 l \choose (5^k 2 l-1)}10 +1-1 \\ =10^{5^k 2 l}-{5^k 2 l \choose 1}10^{(5^k 2l-1)}+ \cdots +(5^k 2 l*10)$

It can have maximum of $5^{k+1}$ as a factor But the dividend is $5^{5^k m}$ .So, obviously $5^k m>(k+1)$ .

So ,there is no $n$