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so I know I need to prove this via induction, but I am somewhat stuck. Here is what I have does so far.

Let $p(n) = (n+1)!^n \le 2!\cdot4!\cdot\ldots\cdot(2n)!$

  1. $p(2) = 3!^2\le 2!\cdot4!$
  2. Assume $p(n)$ is true.
  3. Prove $p(n+1)$
  4. $p(n+1) = (n+2)!^{n+1}\le 2!\cdot4!\cdot\ldots\cdot(2n+2)!$

$$\begin{align*} &=(n+2)!^n\cdot(n+2)!\le\ldots\\ &=(n+1)!^n\cdot(n+2)^n\cdot(n+2)!\le\ldots \end{align*}$$

Given that $(n+1)!^n\le 2!\cdot4!\cdot\ldots\cdot(2n)!$ (Using Inductive Hypothesis)

$$\begin{align*} &=(n+2)^n\cdot(n+2)!\le(2n+2)!\\\\ &=(n+2)^n\le\frac{(2n+2)!}{(n+2)!} \end{align*}$$

And I am stuck here. Any help anyone can give would be helpful.

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For 4, $p(n+1)$ is a statement, not a term, so should not be set equal to something. The $=$ should be logical equivalence. You are trying to prove $p(n) \implies p(n+1)$ This is a nit with your presentation, but is an important concept.

Under 4, you have $(n+1)^n \le \prod_{i=1}^{n}(2i)!$ and want to prove $(n+2)^{n+1} \le \prod_{i=1}^{n+1}(2i)!$. Dividing them we want to prove $\left(\frac {n+2}{n+1} \right)^n(n+2)\le(2n+2)!$ The first paren on the left is less than $e$ so if you have $n$ large enough you are there.

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I'm confused. Which part is wrong? – snowboarder Mar 13 '13 at 3:36
I'm not quite sure how you got 2^(n+1)..etc – snowboarder Mar 13 '13 at 3:46
When you divide the two products how do you get (2n+2) – snowboarder Mar 13 '13 at 4:09
@snowboarder: It is the leftover term-the one that wasn't there in the previous one. Note that the upper limit on the product increased by one, so all the terms in the product from $1$ through $n$ divide out, but the $n+1$ term doesn't. And it has a factorial on it. – Ross Millikan Mar 13 '13 at 4:14
And how about the other side of the equality? How does it become (n+2/n+1)(n+2) – snowboarder Mar 13 '13 at 4:15

Since you have reached $ (n+2)^n.(n+2)! < (2n+2)! $ , I will lead you from here.

From this we can observe that there are 'n' number of $(n+2)$ terms on the LHS.

we can easily conclude that:

$(n+2) < (n+3)$ ---> $1st (n+2)$

$(n+2) < (n+4)$ ---> $2nd (n+2)$

.......and so on till

$(n+2) < (n+n+2)$ ---> $nth (n+2)$

So, we can clearly say that $(n+2)^n.(n+2)! < (n+3).(n+4)....(2n+2).(n+2)!$

=>$(n+2)^n.(n+2)! < (2n+2)!$

Hope the answer is clear !

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I'm still slightly confused by your answer. – snowboarder Mar 13 '13 at 16:03
@snowboarder: Can you mention which part of my answer is confusing so that i can give a more detailed explanation to that part. – lsp Mar 14 '13 at 5:10

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