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
 A: $$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$
so 
$$(1000-1)^n=\sum_{k=0}^n\binom{n}{k}1000^k(-1)^{n-k}$$
all terms except the last is divisible by $1000$, thus $(1000-1)^n\equiv(-1)^n\mod5$
Actually, it's clear that $(1000-1)^n\equiv(-1)^n\mod1000$, so the last $3$ digits of $999^n$ are $999$ when $n$ is odd, and $001$ when n is even.
A: $$(1000-1)^{2016}=\\\left(\begin{array}{c}2016\\ 0\end{array}\right)1000^{2016}+\left(\begin{array}{c}2016\\ 1\end{array}\right)1000^{2015}(-1)^{1}+\left(\begin{array}{c}2016\\ 2\end{array}\right)1000^{2014}(-1)^{2}+...+\left(\begin{array}{c}2016\\ 2015\end{array}\right)1000^{1}(-1)^{2015}+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\\to\\\underbrace{\left(\begin{array}{c}2016\\ 0\end{array}\right)1000^{2016}+\left(\begin{array}{c}2016\\ 1\end{array}\right)1000^{2015}(-1)^{1}+...+\left(\begin{array}{c}2016\\ 2015\end{array}\right)1000^{1}(-1)^{2015}}_{5k}+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\5k+\left(\begin{array}{c}2016\\ 2016\end{array}\right)1000^{0}(-1)^{2016}=\\5k+1$$
A: Units digits are pretty simple. 
Establish a pattern.
The units digits of the powers go $9$, $1$, $9$, $1$...
So $999^{2n}$ or any even power ends in a $1$.
If it ends in a $1$, then it is $1$ mod $5$, and the remainder is $\boxed1$
