How to solve this inequation? So the inequation is this $x^{2016}-1<0 $ 
My initial idea was to transform it like this $x^{2016}>x^0$ and then to look four cases:
$1.$ when $x \lt 0\lt 1$
$2.$ when $x \gt 1$
$ 3.$ when $-1\lt x\lt 0$
$4.$ when $x \lt-1$
Is this the proper way to do it?

As $2016$ is even, the sign of $x$ does not matter.
If we solve in the positives,
$$p^{2016}<1\iff p<1.$$
Hence
$$-1<x<1.$$
