Growth Rate

Let n = 2ᵏ (where k ≥ 1 for n ≥ 2 in (n - 2)!) , then
2ⁿ = 2^2ᵏ , n^2.81 = 2^2.81k , n² 2^√log n = 2²ᵏ 2^√k = 2^(2k+√k) , (log n)²º¹º = k²º¹º , 2010ˡᵒᵍⁿ = 2010ᵏ ≈ 2^10.973k
If k is large enough , then 2k+√k < 2.81k < 10.973k < 2ᵏ ,
so the growth rate of n² 2^√log n < n^2.81 < 2010ˡᵒᵍⁿ < 2ⁿ.
Let k = 2ʳ , then k²º¹º = 2²º¹ºʳ , 2^(2k+√k) = 2^(2ʳ⁺¹ + √2ʳ) ,
if r is large enough (Note : 13 < r min < 14) , then 2010r < 2ʳ⁺¹ + √2ʳ ,
so the growth rate of (log n)²º¹º < n² 2^√log n .
(Note : 2¹³ < log n min < 2¹⁴ ⇒ 2^2¹³ < n min < 2^2¹⁴)
For n ≥ 8 , 2ⁿ < (n - 2)! , so the growth rate of 2ⁿ < (n - 2)!
By the result above , six fuctions in increasing order of growth rate as follow :
(log n)²º¹º < n² 2^√log n < n^2.81 < 2010ˡᵒᵍⁿ < 2ⁿ < (n - 2)!

(n-2)! ....