Prove that in any group, an element has the same order of its inverse

Am I on the right track?

This will be a proof by contradiction.

i. Assume has order n. Therefore,ii. Likewise, assume that has order m, with m different from n. Say, . Therefore,iii. Since and
Multiply both sides by

on the right. Then,

By laws of exponents,

By the definition of order, on the right we have that

So then we have,

iv. Since , as defined at step ii., it follows thatv. But the result of iv. contradicts the claim that a has order n, since by definition this means that n is the least positive integer such that holds.

vi. Therefore an element has the same order of its inverse.

Thanks for anyone who helps! (Itwasntme)