<s>[INST] 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.) 
 [/INST] 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 $9$.
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