Prove that $2^{3^n} + 1$ is divisible by 9, for $n\ge1$ 
Prove that $2^{3^n} + 1$  can be divided by $9$  for $n\ge 1$.

Work of OP: The thing is I have no idea, everything I tried ended up on nothing.
Third party commentary: Standard ideas to attack such problems include induction and congruence arithmetic. (The answers will illustrate, among others, that in this case both approaches work nicely.) 
 A: First, show that this is true for $n=1$:


*

*$\dfrac{2^{3^{1}}+1}{9}=1\in\mathbb{N}$


Second, assume that this is true for $n$:


*

*$\dfrac{2^{3^{n}}+1}{9}=k\in\mathbb{N}$


Third, prove that this is true for $n+1$:


*

*$\dfrac{2^{3^{n+1}}+1}{9}=\dfrac{2^{3^{n}}\cdot2^{3^{n}}\cdot2^{3^{n}}+1}{9}$

*$\dfrac{2^{3^{n}}\cdot2^{3^{n}}\cdot2^{3^{n}}+1}{9}=\dfrac{(9k-1)(9k-1)(9k-1)+1}{9}$ assumption used here

*$\dfrac{(9k-1)(9k-1)(9k-1)+1}{9}=81k^3-27k^2+3k\in\mathbb{N}$
A: Note that 


*

*$2^3$ is $-1$ modulo $9$ 

*$2^{3^n} = (2^{3})^{3^{n-1}}$ 

*$3^{n-1}$ is odd.

A: Hint $\rm \,\ {\rm mod}\,\ A^{\large B} + 1\!:\,\ \color{#c00}{A^B\equiv -1}\ \Rightarrow\ \color{}{A}^{\large BC}\equiv (\color{#c00}{A^{\large B}})^{\large C}\equiv  (\color{#c00}{-1})^{\large C} $   
A: Write $m={3^n}$. Using the binomial theorem, we get $$2^{3^n} + 1=2^m + 1=(3-1)^m+1=9a+3m-1+1=9a+3m$$ which is a multiple of $9$ because $m$ is a multiple of $3$.
A: Now, since we are trying to prove something for all n, we should look immediately to proof by induction.  For $n = 1$, the result is trivial as $2^{3^1}+1 = 9$.  Next, we assume that $2^{3^{(n-1)}} + 1$ is divisible by $9$, say $2^{3^{(n-1)}} + 1=9\cdot k$.  So, $2^{3^{(n-1)}}=9\cdot k -1$.   Now, we examine $2^{3^n}+ 1$ which can be written as $2^{3^{n-1}\cdot3}+1$ which can be written as ${(2^{3^{n-1}}})^3+1$.  Now, since $2^{3^{(n-1)}} = 9 \cdot k -1$ by our induction hypothesis, we have that ${(2^{3^{n-1}}})^3 + 1={(9 \cdot k - 1})^3 +1 = (729\cdot {k^3}- 243\cdot k^2 +27 \cdot k -1)+1 = 729\cdot {k^3}- 243\cdot k^2 +27 \cdot k = 9\cdot (81 k^3 -27 k^2 +3k)$ and so it is divisible by $9$.
