O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

Answer 1

For each constant k asymptotically log(n)^k < n.

Proof is simple, do log on both sides of the equation, and you get:

log(log(n))*k < log(n)

It is easy to see that asymptotically, this is correct.

---

Semantic note: Assuming here log(n)^k == log(n) * log(n) * ... * log(n) (k times) and NOT log(log(log(...log(n)))..) (k times) as it is sometimes also used.