<s>[INST] 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.
 [/INST] 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 $$2016^{2016}\equiv6176\cdot1841\cdot9601\equiv\color{red}{3616}\pmod{ 10^4} $$
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