I have to prove $n! < n^n$ for positive integers greater than 1, but with a little twist. I have to show $P(n-1)$ holds. For the left, I know $(n-1)! = \frac{n!}{n}$ and I'm stuck from there thinking about $(n-1)^{n-1}$.
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I assume "I have to show $P(n-1)$ holds" (whichi will fill for $n=2$) is intendeed to mean "I have to use induction"? – Hagen von Eitzen Nov 15 '13 at 07:35
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So from reading what is below, now I have: n!/n < (n^n)/n < ((n^n)/n) < ((n-1)^n)/(n-1) = (n-1)^(n-1). Did I do this correctly now? – John Hunter Nov 15 '13 at 14:55
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Just to link to the node of this very common inquiry. Actually this post here is older, but the other one has tons of established links already. – Lee David Chung Lin Feb 05 '20 at 12:28
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The inductive proof is standard and will surely be supplied by another answer. Here's a cooler way to do it:
Recall that $n! = \underbrace{n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 2 \cdot 1}_{\text{n elements}}$.
Recall that $n^n = \underbrace{n \cdot n \cdot n \cdot \ldots \cdot n}_{\text{n elements}}$
So both $n!$ and $n^n$ are sequences of length $n$: for the first element of both, $n=n$. Thereafter, the element from $n^n$ is always larger, because $n > n-1, \ldots, n >1$. We know from the axioms of arithmetic and order for $\mathbb{Z}$ that if $a>b$ and $c>0$, then $ac > ab$. So the product of the elements of $n^n$ will be larger.
Newb
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HINT:
If $ m^m> m!,$
$ (m+1)^{m+1}=(m+1) (m+1)^m> (m+1)m^m$ for $m>0$
$\implies (m+1)^{m+1} >(m+1)\cdot m!=(m+1)!$
lab bhattacharjee
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Note that $$n^n=(n-1)^{n-1}\cdot \underbrace{\left(1+\frac1{n-1}\right)^{n-1}}_{>1\text{ if }n\ge 2}\cdot n$$
Hagen von Eitzen
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$1 \lt n$
$2 \lt n$
$3 \lt n$
...
$n-1 \lt n$
$n\le n$
If you multiply you get:
$1\cdot 2\cdot 3\cdot...\cdot (n-1)\cdot n \lt n\cdot n\cdot n\cdot...\cdot n\cdot n$
$n!\lt n^n$
learner
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Thanos Darkadakis
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Would it be acceptable to start from an obviously true statement like $$n! < (n-1)^n + (n-1)!$$ and work backwards to your desired inequality. $$n! < (n-1)^n + (n-1)!$$ $$n!-(n-1)! < (n-1)^n$$ $$\frac{n(n!)-n!}{n} < (n-1)^n$$ $$\frac{(n!)(n-1)}{n}<(n-1)^n$$ $$\frac{n!}{n}<\frac{(n-1)^n}{n-1}$$ $$(n-1)!<(n-1)^{n-1}$$
Aaron Taylor
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I'll leave this here since I've already written it, but Newb has a much clearer and more interesting way of viewing the problem. – Aaron Taylor Nov 15 '13 at 07:51
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If you use induction, the inductive step is
$$(n+1)n!<(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}$$
meta_warrior
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