From what I know, n^n grows faster than n!, so log(n!) is O(log(n^n)) is easy, but how is log(n!) Ω(log(n^n))?
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2[Does this help?](https://stackoverflow.com/a/8118257/5684257) – HTNW Feb 05 '20 at 05:35
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@HTNW definitely! thanks – Joshua Leung Feb 05 '20 at 05:42
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1Does this answer your question? [What is O(log(n!)) and O(n!) and Stirling Approximation](https://stackoverflow.com/questions/8118221/what-is-ologn-and-on-and-stirling-approximation) – walnut Feb 07 '20 at 00:56