Is 2 always a primitive root of 3ˣ? That is, is it always that $$2^{3^x}\equiv -1\pmod{3^{x+1}}\large?$$
 A: Euler function: $\varphi$ 
$\varphi(3^{x+1})=2\cdot 3^x\enspace$ => $\enspace 2^{2\cdot 3^x}\equiv 1 \mod 3^{x+1}\enspace$ (Euler-Fermat)
It follows $\,2^{3^x}\equiv \pm 1 \mod 3^{x+1}$ . 
This means $(2^{3^x}-1)(2^{3^x}+1)\equiv 0\mod 3^{x+1}$ .
$2^{3^x}-1=(3-1)^{3^x}-1\equiv -2 \mod 3$ 
(If this is not clear please have a look to the comment of User user1952009 below.) 
It follows that $2^{3^x}-1$ can never devided by $3$.
$=> \enspace$ $2^{3^x}+1\equiv 0\mod 3^{x+1}\enspace$ which has to be proofed 
A: Another way to solve this question is by induction.
►The statement holds for $n=1$ because $2^3=8=9-1\equiv -1\pmod {3^2}$.
►Suppose it is true for $n$, that is $2^{3^n}\equiv-1\pmod{3^{n+1}}$.
►Proof it is true for $n+1$.
$$2^{3^n}\equiv-1\pmod{3^{n+1}}\iff2^{3^n}=3^{n+1}M_n-1$$ It follows 
$$2^{3^{n+1}}=(2^{3^n})^3=(3^{n+1}M_n-1)^3=3^{3n+3}M_n^3-3\cdot3^{2n+2}M_n^2+3\cdot3^{n+1}M_n-1$$ Hence
$$2^{3^{n+1}}=3^{n+2}(3^{2n+1}M_n^3-3^{n+1}M_n^2+M_n)-1\Rightarrow\color{red}{2^{3^{n+1}}\equiv-1\pmod{3^{n+2}}}$$
