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How can I compare $(\log^*n)!$ with $(n\log n)^b$? I know that $n^b<n!$.

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  • $\begingroup$ Your question has nothing to do with algorithms per se. $\endgroup$ – Yuval Filmus Jan 14 at 7:55
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For sufficiently large values of $n$, and $b>0$:

$$ ( \log^*n )! < ( \log \log n )! < (\log \log n)^{\log \log n} = 2^{(\log\log n) \log \log \log n} \in o(2^{b \log n}) \subset o( (n \log n)^b ). $$

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