Abstract Algebra Proof (Cyclic cycles & order)

[rtw528]

Prove that if G is a group and aεG, then o(a^-1)=o(a)

This is all I have so far:
Assume G is a group and aεG. Because G is a group a has an inverse in the group, a^-1 s.t. aa^-1=e, which is also in G.
<a>={aⁿ|nεZ}. |<a>| is the number of elements in <a> before it cycles back.

Basically all I've done is write what I know about what is given. I have tried to find <a> and <a^-1> from a previous problem to see if there is a pattern but I don't see one.

 

[R136a1]

Can you prove $<a> = <a^{-1}>$?

 

[Dick]

Or can you show $(a^n)^{-1}=(a^{-1})^n$ and use that to show $a^n=e$ if and only if $(a^{-1})^n=e$?