elementary number theory - 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.