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prove that (n!)^2 > n^n

Note by Mishti Angel 5 years, 1 month ago

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It is false for n=1

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Its false for n=2 also

proof: let's take n=1 n! = 1! = 1 (n!)^2 = 1^2 = 1 n^n = 1^1 = 1 then 1 > 1 is false so this expression is false. /* for n=2 n! = 2 n^2 = 2^2 = 4 n^2 = 2^2 = 4 4 > 4 is false

It's true for n = 3, 4, 5, ...

Maybe Mishit think \((n!)^2\geq n^n\)

the question tht i meant is posted by arber nd the inequality holds good for n belonging to natural numbers other than 1 and 2.... we need to prove it theoritically

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestIt is false for n=1

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Its false for n=2 also

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proof: let's take n=1 n! = 1! = 1 (n!)^2 = 1^2 = 1 n^n = 1^1 = 1 then 1 > 1 is false so this expression is false. /* for n=2 n! = 2 n^2 = 2^2 = 4 n^2 = 2^2 = 4 4 > 4 is false

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It's true for n = 3, 4, 5, ...

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Maybe Mishit think \((n!)^2\geq n^n\)

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the question tht i meant is posted by arber nd the inequality holds good for n belonging to natural numbers other than 1 and 2.... we need to prove it theoritically

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