Hurkyl, your genuis quite disgusts me. In any event, I was able to complete the proof. Thanks so much for your help! You weren't by any chance born knowing this, were you?
In case you're interested, you have to instead compute (ab)^q=e^q. Once you get b^(qm)=e, you use the fact that b has order n and that n,m are relatively prime to show nq. Similarly, you can show mq. And so q is a common multiple of m,n, and so mnq.
