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 others have said, a far simpler method would be to just do the following:
$x^{2016}-1<0 \Rightarrow x^{2016}<1 \Rightarrow \ln|x^{2016}|<\ln|1|\Rightarrow\ln|x^{2016}|<0\Rightarrow \ln|x|<0\Rightarrow |x|<1$ 
