Find the last $4$ digits of $2016^{2016}$ 
Find the last $4$ digits of $2016^{2016}$.

Technically I was able to solve this question by I used Wolfram Alpha and modular arithmetic so the worst I had to do was raise a $4$ digit number to the $9$th power. I would do $2016^2 \equiv 4256^2 \equiv \cdots$
 and then continue using the prime factorization of $2016$. I am wondering if there is a better way to solve this.
 A: $$\begin{align}(2000+16)^{2016}&\equiv 2016\cdot 2000\cdot 16^{2015} + 16^{2016}&\pmod{10000}\\
&\equiv 2000+16^{2016}&\pmod{10000}
\end{align}$$
So you need to find the last four digits of $16^{2016}$ and add $2000$.
You can actually reduce the exponent modulo $500$, because $16^{2016}\equiv 16^{16}\pmod{625}$ and $16^{2016}\equiv 0\equiv 16^{16}\pmod{16}$.
So you need to compute $16^{16}\mod{10000}$. This can be done by repeated squaring:
$$16^2=256\\
16^4=256^2\equiv 5536\pmod{10000}\\
16^8\equiv 5536^2\equiv 7296\pmod{10000}\\
16^{16}\equiv 7276^2\equiv 1616\pmod{10000}\\
$$
So the result is $2000+1616=3616$.
A: Since $2016$ is a multiple of $16$, we shall look at the $2016^ {2016}$ modular $625$. 
Well, following from Euler`s Theorem, $2016^{2016} \equiv 2016^{16} \pmod {625}$. Note that $2016^ {15}=(2000+16)^ {15} \equiv16^ {15} \pmod {625}$, following from the Binomial theorem. However, $16^ {15}=2^ {60} \equiv 399^6 \equiv (400-1)^6$, which calculated through the binomial theorem, is $101$. This gives us that $2016^ {16} \equiv 101*2016 \equiv 491 \pmod {625}$. 
Combining this with the upper result gives us $3616$. 
A: Allow me give a different answer from the three ones above and easy to achieve with a calculator, please. This direct procedure can be applied to similar calculations.  
One can simply use one or both properties, $A^{abc}=(A^a)^{bc}\equiv(B)^{bc}\pmod{M}$   and $A^n=A^r\cdot A^s\equiv(B)\cdot (C)\pmod{M}$   where $r+s=n$, iterating the process according to convenience of exponents.$$******$$
$2016^{2016}=(2^5\cdot3^2\cdot7)^{2016}=(2^{2^5\cdot3^2\cdot5\cdot7})(3^{2^6\cdot3^2\cdot7})(7^{2^5\cdot3^2\cdot7})$
Calculating separately each of the three factors, we have
$2^{2^5\cdot3^2\cdot5\cdot7}\equiv(8368)^{6\cdot6\cdot 8}\equiv (9024)^{6\cdot 8}\equiv (8976)^8\equiv6176\pmod{ 10^4} $
$3^{2^6\cdot3^2\cdot7}\equiv(3203)^{8\cdot8\cdot3}\equiv(3761)^{8\cdot3}\equiv(8881)^3\equiv1841\pmod{ 10^4} $
$7^{2^5\cdot3^2\cdot7}\equiv(4007)^{4\cdot4\cdot6}\equiv(5649)^{4\cdot4}\equiv(2401)^4\equiv9601\pmod{ 10^4} $
Hence $$2016^{2016}\equiv6176\cdot1841\cdot9601\equiv\color{red}{3616}\pmod{ 10^4} $$
