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Question Number 69594 by ahmadshahhimat775@gmail.com last updated on 25/Sep/19

Answered by MJS last updated on 25/Sep/19

$$2⩽n⩽4:2^{n!}<2^n!$$$$5⩽n:2^{n!}>2^n!$$$$\ln 2^{n!}=n!\ln 2\Rightarrow \mathrm{adding}n!\mathrm{numbers}\ln 2$$$$\ln (2^{\mathrm{n}}!)=\underset{k=1}{\sum ^{2^n}}\ln k\Rightarrow \mathrm{adding}{2}^n\mathrm{numbers}⩽n\ln 2$$$$n!\ln 2\stackrel{?}{<>}{2}^{n}n\ln 2$$$$n!\stackrel{?}{<>}{2}^{n}n$$$$n⩾6\Rightarrow n!>2^{n}n\Rightarrow ...\Rightarrow 2^{100!}>2^{100}!$$

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