# Comparison of $(\log^* n)!$ and $(n\log n)^b$ [duplicate]

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

• Your question has nothing to do with algorithms per se. – Yuval Filmus Jan 14 at 7:55

## 1 Answer

For sufficiently large values of $$n$$, and $$b>0$$:

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