Solve congruence Solve:$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod {2016}$$
So $ 2016 = 2^5 \cdot 3^2 \cdot 7$
And $$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2^5} \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{3^2} \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod 7 \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
What now?

The lambda-function : https://en.wikipedia.org/wiki/Carmichael_function
helps us to find a quick answer. Because of $\lambda(2016)=24$ and considering $2016=2^5\cdot 3^2\cdot 7$, we can conclude $a^k=a^{k+m\cdot 24}$ for all $a$ and $m$, when $k\ge 5$.
The given number can be written by $2^M$, where $M$ is a power tower of $2's$ with height $2015$. Obviously , $M$ is divisble by $8$ and the remainder modulo $3$ is $1$. 
Applying the chinese remainde theorem , we get $M=16$ modulo $24$. 
So, we can conclude $2^M=2^{16}=1024$ modulo $2016$. 
Another approach :
The given number is obviously divisble by $2^5$. With induction, you can prove that $2^{2^m}=7$ modulo $9$ for every even $m\ge 2$.
The base case $m=2$ is easy to verify.
Then , $2^{2^{m+2}}=(2^{2^m})^4=7^4=7$ mod $9$ is the induction step.
Finally, you can show $2^{2^m}=2$ modulo $7$ for every even $m\ge 2$ analogue to the proof for $9$ (I will leave it to you as an exercise).
So, your number is congruent 
$0$ modulo $32$
$7$ modulo $9$
$2$ modulo $7$
Now, apply the chinese remainder theorem.
