Timeline for Proof for $(n!)^2 \geq n^n$ without using induction [duplicate]

13 events

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May 23, 2021 at 15:59	vote	accept	Abhishek A Udupa
May 23, 2021 at 15:51	history	closed	Martin R Mark Viola vitamin d Henry CommunityBot		Duplicate of Prove that $n! > \sqrt{n^n}, n \geq 3$, Show that if $n>2$, then $(n!)^2>n^n$.
May 23, 2021 at 15:47	comment	added	Abhishek A Udupa		I've edited the question to the right proposition. Also I'd like to prove this without using induction.
May 23, 2021 at 15:46	comment	added	Henry		Does this answer your question? Prove that $n! \gt \sqrt{n^n}, n \geq 3$
May 23, 2021 at 15:45	history	edited	Abhishek A Udupa	CC BY-SA 4.0	added 3 characters in body; edited title
May 23, 2021 at 15:43	answer	added	Ritam_Dasgupta		timeline score: 0
May 23, 2021 at 15:41	answer	added	Thomas Andrews		timeline score: 2
May 23, 2021 at 15:40	comment	added	Martin R		Another one: math.stackexchange.com/q/2919901/42969.
May 23, 2021 at 15:38	comment	added	Martin R		Does this answer your question? Show that if $n>2$, then $(n!)^2>n^n$.
May 23, 2021 at 15:34	comment	added	plop		After you fix the typo, $\frac{((n+1)!)^2/(n+1)^{n+1}}{(n!)^2/n^2}=(n+1)(1+1/n)^{-n}\geq (n+1)e^{-1}$. So, for $n>2$ one term of $\frac{(n!)^2}{n^n}$ to the next differ by a factor grater than $1$. So, you only need to check the first few values $n=1,2$.
May 23, 2021 at 15:29	comment	added	Mark Viola		@Joe And also $n=2$.
May 23, 2021 at 15:28	comment	added	Joe		What about $n=1$?
May 23, 2021 at 15:27	history	asked	Abhishek A Udupa	CC BY-SA 4.0