Prove that $x^{n-1}=x^{-1}$ (Groups)

Question

Suppose $G$ is a group, $x\in G$ and order$(x)=n$.
I have to show that $x^{-1}=x^{n-1}$

Is it sufficient to just say $x^{n-1}=x^nx^{-1}=1_Gx^{-1}=x^{-1}$?

This is a homework question and I'm not sure whether it needs to be done without assuming you can use index laws with regards to elements of a group.

Any other ways or is this okay?

Answer 1

Yes, that would be correct, assuming that you can use the fact that$$(\forall m,m\in\mathbb Z):x^{m+n}=x^mx^n.$$Or you can say that$$1_G=x^n=x^{n-1}.x$$and that therefore $x^{n-1}=x^{-1}$.

Answer 2

Your proof correct and nice.

A slightly different proof goes as follows: $$ x x^{n-1}=x^n=1_G \implies x^{n-1}=x^{-1} $$ either by multiplying on the left by $x^{-1}$ or by using the uniqueness of inverses.