Question: > 
> 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.

Answer: 
$$\begin{align}(2000+16)^{2016}&\equiv 2016\cdot 2000\cdot 16^{2015} + 16^{2016}&\pmod{10000}\\
&\equiv 2000+16^{2016}&\pmod{10000}
\end{align}$$

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 $2000+1616=3616$.
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 $$2016^{2016}\equiv6176\cdot1841\cdot9601\equiv\color{red}{3616}\pmod{ 10^4} $$