$n! = O(n^{n})$
Now $2^{n}$ $n^{n}$ $n^{logn}$
take log $\Rightarrow$ n nlogn lognlogn
it's clear that among these 3 nlogn has the highest growth rate, Now compare n & lognlogn
|
n |
lognlogn |
n=8 |
8 |
9 |
n=16 |
16 |
16 |
n=32 |
32 |
25 |
So, after n = 16 , lognlogn is overtaking n, that's why lognlogn > n
So, $lognlogn<n<nlogn$ which implies only option D