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Question Number 9790 by tawakalitu last updated on 04/Jan/17
prove by mathematical induction  n! ≤ n^n
$$\mathrm{prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\mathrm{n}!\:\leqslant\:\mathrm{n}^{\mathrm{n}} \\ $$
Commented by FilupSmith last updated on 04/Jan/17
n=1  1!≤1^1     n=2  2!≤2^2   2≤4    (n+1)!≤(n+1)^(n+1)   (n+1)∙n!≤(n+1)^(n+1)   n!≤(n+1)^n   1×2×...×n_(n times) ≤(n+1)×...×(n+1)_(n times)   ∀k∈(1,n), (n+1)>k
$${n}=\mathrm{1} \\ $$$$\mathrm{1}!\leqslant\mathrm{1}^{\mathrm{1}} \\ $$$$ \\ $$$${n}=\mathrm{2} \\ $$$$\mathrm{2}!\leqslant\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{2}\leqslant\mathrm{4} \\ $$$$ \\ $$$$\left({n}+\mathrm{1}\right)!\leqslant\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$$\left({n}+\mathrm{1}\right)\centerdot{n}!\leqslant\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${n}!\leqslant\left({n}+\mathrm{1}\right)^{{n}} \\ $$$$\underset{{n}\:{times}} {\mathrm{1}×\mathrm{2}×…×{n}}\leqslant\underset{{n}\:{times}} {\left({n}+\mathrm{1}\right)×…×\left({n}+\mathrm{1}\right)} \\ $$$$\forall{k}\in\left(\mathrm{1},{n}\right),\:\left({n}+\mathrm{1}\right)>{k} \\ $$
Commented by tawakalitu last updated on 05/Jan/17
Thank you sir. God bless you.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by tawakalitu last updated on 05/Jan/17
Thank you sir. God bless you.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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