Is $2^{3^n}+1$ always divisible by $3^{n+1}$? Because it certainly seems to be the case for all positiv $n$ but how can I prove it?
 A: Proof by induction: For $n=1$ it is true. Now assume $2^{3^n}\equiv -1(\text { mod }3^{n+1})$, then
$$
2^{2\cdot3^n}-2^{3^n}+1\equiv ({-1})^2-(-1)+1\equiv 0 (\text { mod }3)
$$
so $3|2^{2\cdot3^n}-2^{3^n}+1$ and $3^{n+1}|2^{3^n}+1$, so 
$$
3^{n+2}|(2^{3^n}+1)(2^{2\cdot3^n}-2^{3^n}+1)=2^{3^{n+1}}+1
$$
A: Suppose
$$2^{3^n} + 1 \equiv 0 \pmod{3^{n+1}}$$
Then
$$2^{3^n} \equiv -1 + 3^{n+1} a \pmod{3^{n+2}}$$
for some value of $a$. And thus
$$\begin{align}2^{3^{n+1}} &\equiv \left(-1 + 3^{n+1} a\right)^3
\\&\equiv (-1)^3 + 3 \cdot (-1)^2 \left(3^{n+1} a \right) + \ldots
\\&\equiv -1 + 3^{n+2}(\ldots) + \ldots
\\&\equiv -1 \pmod{3^{n+2}} \end{align}$$
In fact, we can show more: that $3^{n+2}$ does not divides $2^{3^n}-1$. Repeating a similar inductive argument, if:
$$ 2^{3^n} \equiv -1 + 3^{n+1} a \pmod{3^{n+3}} $$
for some $a$ such that $a \not\equiv 0 \pmod{3} $, then we ave
$$ 2^{3^{n+1}} \equiv -1 + 3^{n+2} a \pmod{3^{n+3}}$$
and so we know that $3^{n+2}$ divides $2^{3^{n+1}} + 1$, but $3^{n+3}$ does not.
