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.) 

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 $3$.
