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I'm having trouble proving that $$n! \leqslant n^n \, \, \, \,\forall \,n \in \mathbb{Z}^+$$ by mathematical induction. I checked if it worked for $n = 1$ and then supposed that it worked for $n$, to then prove if it worked for $n+1$.

In this last step I tried writing $(n+1)!$ like $n!(n+1)$ but I don't know how to continue. Thank you so much.

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    $\begingroup$ You're on the right track: $n!(n+1) \leqslant n^n(n+1) \leqslant(n+1)^n(n+1)=(n+1)^{n+1}$ $\endgroup$
    – M.P
    Jun 19, 2019 at 9:54
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    $\begingroup$ A simple search can get you many relevant answers for common questions. $\endgroup$
    – Wei Zhong
    Jun 19, 2019 at 9:55
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    $\begingroup$ Here is a video explaining: youtu.be/NsO6nh42oPo $\endgroup$
    – user679128
    Jun 19, 2019 at 9:56
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    $\begingroup$ Another approach: Note that $\frac{n!}{n^n} = \prod_{k=1}^n \frac{k}{n}$. Then, use induction to prove that the product of $n$ numbers less than $1$ is also less than $1$. $\endgroup$ Jun 19, 2019 at 9:59

3 Answers 3

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Let $n! \le n^n$.

Since $n^n \le (n+1)^n$ we get

$$(n+1)!=n!(n+1) \le n^n(n+1) \le (n+1)^n(n+1) = (n+1)^{n+1}.$$

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  • $\begingroup$ Thank you so much! You solved my problem. $\endgroup$ Jun 19, 2019 at 9:55
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$n! (n+1) \le n^n(n+1)$ by induction assumption, and $n^n(n+1) < (n+1)^n(n+1)$ because $n<n+1$.

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$n^{n}<(n+1)^{n}$ So $(n+1)(n^{n})<(n+1)^{n+1}$. Hence, if we assume that $n! <n^{n}$ we get $(n+1)!=(n+1) n! <(n+1) n^{n}<(n+1)^{n+1}$

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