# Thread: Prove x^n > x for all x > 1.

1. Originally Posted by lvleph
Yes, because the hint was to prove $f(x) > g(x)$ and we wanted to prove $x^n > x$ so it makes sense to assume $g(x)=x$.

Factorise! $x^n-x=x(x^{n-1}-1)=x(x-1)(1+x+x^2+\cdots+x^{n-2}),$ the product of three terms which are obviously positive for $x\,>\,1.$