Please see the " Microsoft word" attachment.
$\displaystyle 2^{2^4}=2^{16}=65536\equiv 1\ \mod\ 5$
Also
$\displaystyle
2^{4k}=(2^4)^k \equiv 1 \ \mod \ 5
$
and since the $\displaystyle \log_2$ of all the numbers in the sequence after $\displaystyle 2^{2^2}$ are divisible by $\displaystyle 4$ the result follows for $\displaystyle \mod\ 5$.
By a similar argument we can show that if any element in the sequence is ever congurent to 0 or 1 so
are all subsequent terms.
RonL