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... 
 A: 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.
A: $ \overbrace{2^{\large 2^{\Large 2K}}\!\!\!\bmod 2^{\large 5}\!\cdot 63}^{\large\ \ \   2^{\Large 2K}\ge5\ \ {\rm by}\ \ K>1 }\, =\, 2^{\large 5}\!\left[\dfrac{{2^{\large \color{#c00}{2^{\Large 2K}}}}}{2^{\large 5}} \bmod\, 63\right] =\, 2^{\large 5}\overbrace{ \left[\,\dfrac{{2^{\large \color{#c00}{4}}}}{2^{\large 5}} \bmod 63\right]}^{\!\!\!\! \dfrac{2^{\large 5}}{2^{\large 6}_{\phantom{1}}}\ \ {\large \equiv}\ \ \dfrac{2^{\large 5}}{1}} =\, \bbox[5px,border:1px solid #c00]{2^{\large 5}[\,2^{\large 5}\,]}\ \ $  by
$\!\!\bmod 63\!:\ 2^{\large\color{#0a0} 6}\!\equiv 1\,$  so $\!\underbrace{\color{#c00}{2^{\large 2K}}\!\bmod\color{#0a0} 6_{\phantom{1}}}_{\large 2\ \mid\ 2^{\Large 2K}\  {\rm by} \ K>1}\!\!\! = 2\!\!\!\underbrace{\left[\dfrac{2^{\large 2K}}{2}\!\bmod 3\right]}_{  \dfrac{(-1)^{2K}}{-1}\ {\large \equiv}\ \dfrac{1}{-1} {\large}\ {\large \equiv}\ \  \large  2}\!\!\!\!\! =\color{#c00} 4$
