Is (lg n)! in O((lg n)^(lg n))?

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Dragonfall
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[tex](\lg n)!\in\mathcal{O}((\lg n)^{\lg n})[/tex] right?
 

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Right, you can get that from Stirling's formula. (Assuming that either n is a power of 2 or you use gamma(lg(n) + 1), of course.)
 
Excellent, thanks.