Suppose that n does divide . Let p be the smallest prime divisor of n, then p must divide . In other words, . Since p is prime, it follows from Fermat's Little Theorem that . Then by part (i), , contradiction!
Edit. I forgot to say that that proof doesn't work if p=2 (because 2 and p are not then co-prime, so Fermat's theorem goes wrong!). But the result is obvious when p=2, because n would then be even, and (2^n)-1 is odd.