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.) 

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 $9$.
