Simplifying factorials

Could someone kindly explain to me the passages underlying the result of the following division... How does

(n+2)^n\times(n+2)!\leq (2n+2)!

when divided by (2n+2)! become

\frac{n+2}{2n+2}\times\frac{n+2}{2n+1}\times...\times\frac{n+2}{n+3} \leq 1

I am concerned about the left side of the inequality.

Thank you very much.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

Hint: Show that $\dfrac{(2n+2)!}{(n+2)!} = (2n+2)(2n+1) \cdots (n+3)$ .

Pi Han Goh - 3 years, 4 months ago

Dear Pi Han Goh... thank you for your reply. I have tried, but am not able. I am just beginning to learn about these types of expressions. Could you show me how you would do it?

Thank you again.

Eugenia Fiammetta Rossetti - 3 years, 4 months ago

Actually I got it...!!!! Thank you!!!!!!!

Eugenia Fiammetta Rossetti - 3 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$