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2-100-2-100-lt-or-gt-

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Question Number 225970 by fantastic2 last updated on 17/Nov/25
2^(100!) ? 2^(100) ![=,<or >]
$$\mathrm{2}^{\mathrm{100}!} \:?\:\mathrm{2}^{\mathrm{100}} !\left[=,<{or}\:>\right] \\ $$
Commented by Ghisom_ last updated on 18/Nov/25
b^(n!) vs. b^n ! where b, n ∈N∧b≥2 for huge n we can use n!≈((n/e))^n (√(2πn)) ⇒ log_b b^(n!) ≈((n/e))^n (√(2πn)) log_b b^n ! ≈ b^n (1−(1/(ln b)))+((ln 2π)/(2ln b))+(n/2) ⇒ there′s an m∈N: n>m ⇒ b^(n!) >b^n ! for b=2 we have m=4. for k≥1 2^((4+k)!) >2^(4+k) ! ⇒ 2^(100!) ≫2^(100) ! ≈10^(2.8×10^(157) ) ≫10^(3.8×10^(31) )
$${b}^{{n}!} \:\mathrm{vs}.\:{b}^{{n}} !\:\:\mathrm{where}\:{b},\:{n}\:\in\mathbb{N}\wedge{b}\geqslant\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{huge}\:{n}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:{n}!\approx\left(\frac{{n}}{\mathrm{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}} \\ $$$$\Rightarrow \\ $$$$\mathrm{log}_{{b}} \:{b}^{{n}!} \:\approx\left(\frac{{n}}{\mathrm{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}} \\ $$$$\mathrm{log}_{{b}} \:{b}^{{n}} !\:\approx\:{b}^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{ln}\:{b}}\right)+\frac{\mathrm{ln}\:\mathrm{2}\pi}{\mathrm{2ln}\:{b}}+\frac{{n}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:{m}\in\mathbb{N}:\:{n}>{m}\:\Rightarrow\:{b}^{{n}!} >{b}^{{n}} ! \\ $$$$\mathrm{for}\:{b}=\mathrm{2}\:\mathrm{we}\:\mathrm{have}\:{m}=\mathrm{4}.\:\mathrm{for}\:{k}\geqslant\mathrm{1} \\ $$$$\mathrm{2}^{\left(\mathrm{4}+{k}\right)!} >\mathrm{2}^{\mathrm{4}+{k}} ! \\ $$$$\Rightarrow \\ $$$$\mathrm{2}^{\mathrm{100}!} \gg\mathrm{2}^{\mathrm{100}} ! \\ $$$$\approx\mathrm{10}^{\mathrm{2}.\mathrm{8}×\mathrm{10}^{\mathrm{157}} } \gg\mathrm{10}^{\mathrm{3}.\mathrm{8}×\mathrm{10}^{\mathrm{31}} } \\ $$
Commented by fantastic2 last updated on 18/Nov/25
2^(100) !=1×2×3×4...×(2^(100) −2)(2^(100) −1)(2^(100) )_(2^(100) times) 2^(100) >1 2^(100) >2 ... 2^(100) >(2^(100) −1) 2^(100) =2^(100) so 2^(100) !<(2^(100) )^2^(100) 2^(100!) =(2^(100) )^(99!) now 99!≫2^(100) so 2^(100!) ≫2^(100) !
$$\mathrm{2}^{\mathrm{100}} !=\underset{\mathrm{2}^{\mathrm{100}} \:{times}} {\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}…×\left(\mathrm{2}^{\mathrm{100}} −\mathrm{2}\right)\left(\mathrm{2}^{\mathrm{100}} −\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{100}} \right)} \\ $$$$\mathrm{2}^{\mathrm{100}} >\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{100}} >\mathrm{2} \\ $$$$… \\ $$$$\mathrm{2}^{\mathrm{100}} >\left(\mathrm{2}^{\mathrm{100}} −\mathrm{1}\right) \\ $$$$\mathrm{2}^{\mathrm{100}} =\mathrm{2}^{\mathrm{100}} \\ $$$${so}\: \\ $$$$\mathrm{2}^{\mathrm{100}} !<\left(\mathrm{2}^{\mathrm{100}} \right)^{\mathrm{2}^{\mathrm{100}} } \\ $$$$\mathrm{2}^{\mathrm{100}!} =\left(\mathrm{2}^{\mathrm{100}} \right)^{\mathrm{99}!} \\ $$$${now}\:\mathrm{99}!\gg\mathrm{2}^{\mathrm{100}} \\ $$$${so}\:\mathrm{2}^{\mathrm{100}!} \gg\mathrm{2}^{\mathrm{100}} ! \\ $$

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