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.) 

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} $   
