First, the problem says to arrange the functions in increasing, not decreasing, order. Second, it is not clear whether you mean (2^100)n or 2^(100n), 2n! or 2^(n!). The function n^log n = O(2^(100n)) and n^(4/3) (log n)^2 = O(n^log n).I tried the problem out. 2n!, 2^2^n,2^n^2, 2^100n, n^4/3 (logn)^2, n^3/2, n(logn)^3/2, nlogn loglogn. How far off am I?