Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$. Prove or disprove that there isn't any positive integer $n>7$ 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.
