Calculate $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 2016$ How to find $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 2016$ where $2$ occurs $2016$ times? 

My current observations:
$$2^{11} = 2048 \equiv 2048=2016 \equiv 2^5 $$
and
$$ 2^{16} \equiv 2^{11}\cdot 2^5 \equiv 2^{10} $$ and now we have $2012$ of "2" left... 

Hint: $2016 = 2^5 \cdot 3^2 \cdot 7$.  Consider it separately mod $2^5$, mod $3^2$ and mod $7$, and combine using the Chinese Remainder Theorem.
