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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 ⇒ adding n! numbers ln 2 ln (2^n !) =Σ_(k=1) ^2^n ln k ⇒ adding 2^n numbers ≤ n ln 2 n! ln 2 <>^(?) 2^n n ln 2 n! <>^(?) 2^n n n≥6 ⇒ n!>2^n n ⇒ ... ⇒ 2^(100!) >2^(100) !

$$\mathrm{2}\leqslant{n}\leqslant\mathrm{4}:\:\mathrm{2}^{{n}!} <\mathrm{2}^{{n}} ! \\ $$$$\mathrm{5}\leqslant{n}:\:\mathrm{2}^{{n}!} >\mathrm{2}^{{n}} ! \\ $$$$\mathrm{ln}\:\mathrm{2}^{{n}!} \:={n}!\:\mathrm{ln}\:\mathrm{2}\:\Rightarrow\:\mathrm{adding}\:{n}!\:\mathrm{numbers}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{ln}\:\left(\mathrm{2}^{\mathrm{n}} !\right)\:=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} } {\sum}}\mathrm{ln}\:{k}\:\Rightarrow\:\mathrm{adding}\:\mathrm{2}^{{n}} \:\mathrm{numbers}\:\leqslant\:{n}\:\mathrm{ln}\:\mathrm{2} \\ $$$${n}!\:\mathrm{ln}\:\mathrm{2}\:\overset{?} {<>}\:\mathrm{2}^{{n}} {n}\:\mathrm{ln}\:\mathrm{2} \\ $$$${n}!\:\overset{?} {<>}\:\mathrm{2}^{{n}} {n} \\ $$$${n}\geqslant\mathrm{6}\:\Rightarrow\:{n}!>\mathrm{2}^{{n}} {n}\:\Rightarrow\:...\:\Rightarrow\:\mathrm{2}^{\mathrm{100}!} >\mathrm{2}^{\mathrm{100}} ! \\ $$

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