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: 
You can afford to work modulo $16=2^4$ and modulo $625=5^4$ separately. The basis $2016$ is a multiple of $16$ and therefore so are all its powers.

We also know that $2016\equiv141$ modulo $625.$

To take the $2016$-th power note that $141$ is invertible modulo $625$ (because it is relatively prime with $5$) and therefore its $500$-th power is equivalent to $1$ modulo $625$ (the multiplicative group modulo $625$ has order $5^3.(5-1)=500$). Thus we can limit ourselves to the $16$-th power.

To evaluate $141^{16}$ modulo $625$ notice that $141=5^3+16$ so

$$\eqalign{141^{16}=\sum\_{i=0}^{16}{16\choose i}5^{3i}16^{16-i}
&\equiv16^{16}+16.16^{15}.125\\
&\equiv16^{16}+1.125\\
16^{16}=(3.5+1)^{16}=\sum\_{i=0}^{16}{16\choose i}3^i5^i
&\equiv 1+16.3.5+8.15.3^2.5^2+8.5.14.3^3.5^3\\
&\equiv1+48.5+216.125\\
&\equiv1+48.5+1.125=366\\
141^{16}&\equiv366+125=491\\
}$$

This means we are looking for a number between $0$ and $9999$ that is a multiple of $16$ and that is $491$ more than a multiple of $625.$ Since $625\equiv1$ and $491\equiv11$ modulo 16 the solution should be $(16-11).625+491=3616.$
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 $$2016^{2016}\equiv6176\cdot1841\cdot9601\equiv\color{red}{3616}\pmod{ 10^4} $$