Formally, you also need to consider the case when m > n, but I agree with your proof so far.
It may be easier to just note that for all integer n. Therefore, iff . So, . From here the claim follows.
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 andiv. Since , as defined at step ii., it follows that
Multiply both sides by on the right. Then,
By laws of exponents,
By the definition of order, on the right we have thatSo then we have,
v. 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!