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#### issacnewton

##### Member

- Jan 30, 2012

- 61

I wish to prove that

\[ \forall\;n\in \mathbb{N}\; n! \leqslant n^n \]

First we let \(n\) be arbitrary. Now I first write \( n! \) as \( n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1\).

Now we see that

\[ n \geqslant (n-1)\;; n \geqslant (n-2)\;\ldots ;n \geqslant n- (n-1) \]

So we get

\[ n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1 \leqslant \underbrace{n\cdot n\cdot n\cdots n}_\text{n times} \]

\[ \Rightarrow n! \leqslant n^n \]

Since \(n\) is arbitrary, the result is generally true. I want to use this result to find the limit of \( (n!)^{1/n^2 } \)

using the Squeeze theorem. So is my proof correct ?

Thanks