Invertible element can't be nilpotent?

Question

When I was reviewing cracking GRE subject mathematics 4-th edition, I was confused about the proof in Page 247.

Consider invertible element c in a Ring, $cc^{-1} = 1$. Then for any integer n, $(cc^{-1})^{n} = 1$, which means $c^{n}c^{-n} = 1$. Then $c^{n}$ can't be zero for any n, which means c can't be nilpotent.

I was wondering if $(cc^{-1})^{n} = c^{n}c^{-n}$ is satisfied for any Ring? I think it's satisfied only if it comes to a commutative ring (multiplication is commutative).

Any help would be greatly appreciated!

Answer 1

Both $(cc^{-1})^n = (c c^{-1}) (c c^{-1}) \cdots (c c^{-1}) = 1$ and $c^n c^{-n} = (c c \cdots c) (c^{-1} c^{-1} \cdots c^{-1}) = 1$, so naturally they are equal.

Answer 2

In any ring an invertible element commutes with its inverse.

Answer 3

In any ring $R$, if any two elements $a,b\in R$ commutes then the identity $$(ab)^n=a^nb^n$$ holds true. It is not necessary that the whole ring should be commutative. Therefore in your case because $cc^{-1}=1=c^{-1}c$, it implies that $c$ and $c^{-1}$ commutes. So that $c^nc^{-n}=(cc^{-1})^n=1.$