Analysis of algorithm about complexity

Question

$n$ is $O(\log n)^{\log n}$ ? This is true or false, Give the reasons behind that ? I dont get understand about that $O(\log n)^{\log n}$.

Answer 1

The question is, wheter $n$ can be written as $n = f(n)^{\log n}$ where $f$ is a function such that $f\in O(\log)$. This is true, as $$ n = 2^{\log n}$$ an the constant function $f(n) := 2$ has $f \in O(\log)$ as $\log$ is unbounded.

Answer 2

In addition to martini's answer, it's also useful to write that $n = e^{\log n} = o(\log n ^{\log n})$ because essentially as $n \to \infty \ \frac{f_1(n)}{f_2(n)}$ converges to $0$, so $f_1(n)$ is of a lesser order than $f_2(n)$.