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check my proof, I feel like I made a mistake :)

so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.

Base Case $$ p(2) \iff 2!<2^2 \iff 2<4 $$

Assume p(k) is true $$k!<k^k$$

Prove p(k+1)

$$(k+1)!<(k+1)^{(k+1)}$$ $$(k+1)(k)!<(k+1)(k+1)^k$$ $$(k+1)(k)!<(k+1)(k^k+1)$$ This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$? $$(k+1)(k)!<(k+1)(k^k+1)$$ $$(k)!<k^k+1$$ Then above, can I factor out a k+1 from both sides? $$(k)!<k^k+1$$

Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?

trying again

prove p(k+1)

to start, im now looking to multiply the sides by (k+1)? not replace?

so

$$k!<k^k$$ $$(k+1)k!<(k+1)k^k$$ $$(k+1)!<(k^[k+1]+k^k$$

Martin Sleziak
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Mike
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  • You should start with $(k+1)!<k^k(k+1)$ – Adam Hughes Jul 01 '14 at 03:15
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    Also, $(k+1)^k\neq k^k+1^k$. – vadim123 Jul 01 '14 at 03:17
  • $2^2 \neq 2$ so this should be edited – Squirtle Jul 01 '14 at 03:24
  • When you are proving that $A=B$ or that $A\lt B$, do not ever start from $A=B$ or $A\lt B$ and manipulate. – André Nicolas Jul 01 '14 at 03:24
  • cool thanks for the feedback guys. Adam, could you explain in short real quick why I would start like that? I just thought when I was looking for p(k+1) and that meant to replace all of the k with k+1? – Mike Jul 01 '14 at 03:37
  • andre where would you start from in that general case then? most of this is conceptual stuff i'm still trying to figure out – Mike Jul 01 '14 at 03:38
  • It would be good if the last formula you wrote were $(k + 1)! < (k + 1)^{(k+1)}$. It is what you are trying to show. Instead you introduced it almost at the beginning, as if it were something you had assumed. – David K Jul 01 '14 at 03:39
  • like I said, I just thought I replaced the k's with k+1's, hence I started with that. So really as my first step I'm multiplying everything by k+1? not replacing it? – Mike Jul 01 '14 at 03:41
  • You are trying to use the assumed fact that $k!\lt k^k$ to prove that $(k+1)!\lt (k+1)^{k+1}$. How do we use the fact to get information about $(k+1)!$? Maybe by using $(k+1)!=(k+1)k!$, We have information about the size of $k!$. We should be able to exploit that to get information about the size of $(k+1)k!$, – André Nicolas Jul 01 '14 at 03:42
  • The idea of replacing each $k$ with $k + 1$ is a good one, but it is how you determine the a goal you have to reach. You don't win the World Cup by saying "OK, I say there is a ball in the other team's goal--so my team scores". You actually have to move the ball there from a legitimate starting point (in this case, from $k! < k^k$). – David K Jul 01 '14 at 03:47
  • ok cool you guys are awesome. I have 0 experience with proofs. Basically what I've been trying to do with the proofs is find a singular formula for solving them, but based on what you just said there really isn't a set formula... you're just trying to find a way from the given to the "goal"? so im moving from $k!<k^k$ to $(k+1)!<(k+1)^k+1$, basically using whatever means I can to make the $k^k$ side look like $(k+1)^(k+1)$? (man i suck at the formatting on this site) – Mike Jul 01 '14 at 03:55
  • can anyone give me a quick yes or no to this? its kind of important to my understanding – Mike Jul 01 '14 at 04:54
  • See also: Recursive proof that $n^n \geq n!$ – Martin Sleziak Jan 26 '19 at 14:58

2 Answers2

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Finishing up the induction proof, \begin{align} (k+1)!&=(k+1)k! \\ &<(k+1)k^k & \text{by hypothesis, } k! < k^k\\ &<(k+1)(k+1)^k & k < k+1 \text{ for all }k>1\\ &=(k+1)^{k+1} \end{align}

Cookie
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  • are you essentially only allowed to tack on the +1 to the $k^k$ because of the "<" ? or could this be done if this were not an inequality proof? i guess im having trouble understanding the basis for adding the +1 – Mike Jul 01 '14 at 03:58
  • like you're converting the $k^k$ because $(k+1)^k$ would still be greater than k or something? – Mike Jul 01 '14 at 04:06
  • Yes, you can tack on the "$+1$" because $k < k+1$ whenever $k>1$. For example, if $k=2$, then $2 < 3$. If $k=5000$, then $5000 < 5001$, and so on. – Cookie Jul 01 '14 at 04:27
  • cool, thanks for clarifying that – Mike Jul 01 '14 at 04:35
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As an alternative solution, we could use the AM-GM inequality:

$$\frac{1 + 2 + 3 + \dots + n}{n} \ge \sqrt[n]{n!}$$ $$\frac{n(n+1)}{2n}\ge \sqrt[n]{n!}$$ $$\frac{n+1}{2}\ge\sqrt[n]{n!}$$

Since $n> \frac{n+1}{2}$ for $n>1$, we have

$$n>\sqrt[n]{n!}$$

And finally,

$$n^n > n!$$

Yiyuan Lee
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