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I'm trying to find a proof for the following statement, using mathematical induction:

$$ (\forall n\in \mathbb N-\{0\}) n^n \ge n! $$

But I always get to a dead-end.

I've done the basis step, for $n = 1$, which is clearly true, but I can't prove for an arbitrary $k+1$ if I assume it's true for $k$.

How can I prove this?

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  • $\begingroup$ Divide both sides by $n$, the inequality's truth is preserved. Can you show $n^{n-1}\geq(n-1)^{n-1}$? $\endgroup$ – MickG Oct 17 '15 at 20:01
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You can use $$(k+1)!=(k+1)\cdot k!$$ and $$k^k\le (k+1)^k.$$ as the following :

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

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we have to prove that $(n+1)!\le (n+1)^{n+1}$ now we get $(n+1)!\le (n+1)\cdot n^n$ and it is easy to show that $(n+1)\cdot n^n \le (n+1)^{n+1}$ the last inequality is true since $n\le n+1$

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