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

$$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$
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.
