Take log on both sides of the functions and compare each functions.
eg 2n n!
taking log on both sides
nlog(base 2) log(n!)
and put n = 2^128 or some 2^k values
ll get
nlog(base2) < log(n!)
and simlarly for 2^n and n^logn
now, 2^n > n^logn
so n! > 2^n > n^logn
Answer is (D).
correct me if i'm wrong
2n