Other than using the units digit division method, how would I demonstrate $999^{2016}$ 's remainder when it is divided by 5? 
Using the other than the units digit division method, how would I demonstrate $999^{2016}$ 's remainder when it is divided by 5?

I have decided that the best method in solving the problem was to engage with a binomial theorem as demonstrated below:
$$(1000-1)^{2016}$$
but then from here I didn't know how to proceed and I really would like to use the binomial theorem method

$$(1000-1)^{2016}=\\\left(\begin{array}{c}2016\\ 0\end{array}\right)1000^{2016}+\left(\begin{array}{c}2016\\ 1\end{array}\right)1000^{2015}(-1)^{1}+\left(\begin{array}{c}2016\\ 2\end{array}\right)1000^{2014}(-1)^{2}+...+\left(\begin{array}{c}2016\\ 2015\end{array}\right)1000^{1}(-1)^{2015}+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\\to\\\underbrace{\left(\begin{array}{c}2016\\ 0\end{array}\right)1000^{2016}+\left(\begin{array}{c}2016\\ 1\end{array}\right)1000^{2015}(-1)^{1}+...+\left(\begin{array}{c}2016\\ 2015\end{array}\right)1000^{1}(-1)^{2015}}_{5k}+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\5k+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\5k+1$$
