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By the inequality here or using integration I can get the following lower bound for $\log(n!)$. How can I get a better lower bound for $\log(n!)$?

$$\log(n!)>\log\left(\frac{(n+1)^n}{e^n}\right)$$

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    The log's in your question are superfluous. You are essentially lower-bounding $n!$. There are very good lower-bounds, notably Sterling's approximation gives $n! > \sqrt{2\pi n}(n/e)^n$.2011-03-27
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    I think you can develop that bound by looking at the Euler-McLaurin summation of $\sum_{k=1}^n \log k$.2011-03-27
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    Re Fixee's comment, see http://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates.2011-03-27

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