Discuss the growth of n^(loglogn)! lim n approaches infinity of f(n)/g. If its 0, calculate the limit one more time. if its infinity or a constant stop. Basically you find the limit of n^(loglong)! divided by a^n. Assume log are base 2, so you can do 2^n. If its 0, then you find the limit using n^k, then n^a, log n, and c until you get infinity or a constant. It will be either big omega, or big theta.