O(n*log(n)) = O(log(n!))?

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Possible Duplicate:
Is log(n!) = Θ(n·log(n))?

Wikipedia states that O(n*log(n)) = O(log(n!)). I don't quite get why. Can anybody explain or prove why it is that way?

possible duplicate of [Is log(n!) = Θ(n·log(n))?](http://stackoverflow.com/questions/2095395/is-logn-nlogn) and [Lower bound for sorting by comparison](http://stackoverflow.com/questions/7233553/lowed-bound-for-sorting-by-comparison). — Raymond Chen, Oct 21 '11 at 14:18

score 1 · Answer 1 · answered Oct 21 '11 at 14:19

1

Have a look at the page on Stirling's Approximation

answered Oct 21 '11 at 14:19

acm

1 Answers1