Is it true that the order of $ab$ is always equal to the order of $ba$?
Solution 1:
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
Solution 2:
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.