# Comparison of two numbers

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

• This en.wikipedia.org/wiki/Stirling%27s_approximation will help. – Michael Hoppe Dec 9 '13 at 12:13
• 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 9 '13 at 12:13
• – lab bhattacharjee Dec 9 '13 at 18:14

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 $n=300$.
• @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 9 '13 at 12:30
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.