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Could you please check if I correctly compared functions? I used basic theorems about big-O notation:

$$ \ln(\ln(n))<\log_2(n^n)<\ln^2(n)<(\log_2 (n))^n<5n \log_2 (n)<(n+1)^2

$\ln(n)$ is the natural log of $n$.

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    This seems like an un-necessarily long chain if you just mean to compare $\log \log (n) $ with $2^{2^n}$. Take the log of both sides to see that $$\log \log \log (n) << 2^n \log 2$$ especially since $ \log \log \log (n) << n$. – 2017-02-06
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    Well the task is to compare all that 9 functions. Is my chain correct? – 2017-02-06
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    @AndresMejia I am the most worry with the parts with logarithms. – 2017-02-06
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    No. $(\log_2 n)^n = o (\log_2 (n^n)) = o (n \log_2 n)$. – 2017-02-06
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    @zhoraster why $(log_2 n)^n = o(log_2(n^n))$? I agree with the secong equals but I don't understand the first one – 2017-02-06
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    To compare, say, $\log(n^n)$ and $(\log n)^n$, first note $\log(n^n)=n\log n$, and then take the log of each of the two quantites. For the first one, you get $\log n+\log\log n$, for the second, $n\log\log n$. Now it should be easy to see which is bigger. This method works for many of the steps in your chain. – 2017-02-06
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    Thank you very much @GerryMyerson! So $log(n^n)<(log(n))^n$. For the next one if I want to compare $ln^2 (n)$ with $log_2 (n^n)$ and I take log for the both functions I have: $2log( ln n) > log (n log_2 n)$. Is it correct? – 2017-02-06
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    No. $\log(n\log n)=\log n+\log\log n>2\log\log n$. – 2017-02-06

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