I wonder what is greater $100^{300}$ or $300!$ ? And how to prove it? Thanks in advance.
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This http://en.wikipedia.org/wiki/Stirling%27s_approximation will help. – Michael Hoppe Dec 09 '13 at 12:13
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I'd say that the larger is $300!$ because $\log n! \approx n \log n$ and so $\log 300! \approx 300 \log 300 > 300 \log 100 = \log 100^{300}$, but this is not a proof. – lhf Dec 09 '13 at 12:13
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Related : http://math.stackexchange.com/questions/562538/which-is-larger-20-or-240 – lab bhattacharjee Dec 09 '13 at 18:14
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For an exact (and conclusive) answer, I'd take base ten logarithms. The easy one is $\log (100^{300}) = 300\cdot \log 100 = 600$. The more difficult one is $$ \log (300!) = \sum_{i = 1}^{300}\log i $$ which WolframAlpha says is about $614.5$. So $300!$ is the larger one, by 14 digits.
Arthur
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Wikipedia gives this elementary estimate based on integrals: $$ n! \ge e\left(\frac ne\right)^n $$ Since $2<e<3$, this implies $\displaystyle n! > \left(\frac n 3\right)^n$, as required: just take $n=300$.
lhf
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@DietrichBurde, yes, indeed. But it does sound a bit silly to be haggling about a factor of $2$ when $300!>(3\times 10^{14})\cdot 100^{300}$. :-) – lhf Dec 09 '13 at 12:30