Q:

The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.

Prove that for every $n\in \mathbb N$, $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.

I was able to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$ using induction.
First, I check whether the statement is true for $n=1$. The statement holds since:
$$2^{3^1}+1=9$$ $$3^{1+1}=9$$
Second, I assume that the statement is true for every $n\in \mathbb N$.
Finally, I take that $n=n+1$. From that, the original expression becomes $(2^{3^n})^3+1$, which is a sum of cubes, and the same as:
$$(2^{3^n}+1)((2^{3^n})^2-2^{3^n}+1)$$
As I assumed that $2^{3^n}+1$ is divisible by $3^{n+1}$, and one of the factors is $2^{3^n}+1$, the entire expression must be divisible by $3^{n+1}$, and so the statement is true for every $n\in \mathbb N$.
Now, I don't know how to prove that $2^{3^n}+1$ isn't divisible by $3^{n+2}$. Please explain how I can do so step-by-step, without using complicated terms.

A:

Let
$$A_n = 2^{3^n}+1.$$
The claim holds for $n=0$, and since:
$$ A_{n+1} = (A_n-1)^3+1 = A_n^3 -3A_n(A_n-1) $$
we have that $\nu_3(A_{n+1}) = 1+\nu_3(A_n)$, since $9\mid A_n^3$ and $3\mid 3A_n$ but $3\nmid (A_n-1)$.

