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I wonder what is greater $100^{300}$ or $300!$ ? And how to prove it? Thanks in advance.

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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 $n=300$.

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  • $\begingroup$ @DietrichBurde, yes, indeed. But it does sound a bit silly to be haggling about a factor of $2$ when $300!>(3\times 10^{14})\cdot 100^{300}$. :-) $\endgroup$ – lhf Dec 9 '13 at 12:30
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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.

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