<s>[INST] Is 2 always a primitive root of 3ˣ? That is, is it always that $$2^{3^x}\equiv -1\pmod{3^{x+1}}\large?$$
 [/INST] 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
$$2^{3^{n+1}}=3^{n+2}(3^{2n+1}M_n^3-3^{n+1}M_n^2+M_n)-1\Rightarrow\color{red}{2^{3^{n+1}}\equiv-1\pmod{3^{n+2}}}$$
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