Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$. Prove or disprove that there isn't any positive integer $n&gt;7$ such that the linear congruence below is true.

$  2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $
Consider $2^{3^4}$, and it's easy to show $1\equiv2^{3^4}=2^{81} \pmod{511}$. Therefore, take $n=511$, $2^{3^{4^{.^{.^{.^{n-1}}}}}}-1$ is apparently a multiple of $2^{3^4}-1$ since $3^{4^{.^{.^{.^{n-1}}}}}$ is divisible by $3^4$ and 
$$2^{3^{4^{.^{.^{.^{510}}}}}}\equiv1 \pmod{511}$$
holds.
Note that $2$ mod $7$ has order $3$ (since $2^2 \equiv 4 \bmod 7$ and $2^3 \equiv 1 \bmod 7$).

So $2^{3^m} \equiv 1 \bmod 7$ for any $m\geq 1$.