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I would like to show that:

$$ \prod_{i=0}^{n} {n \choose i}\leq \left(\frac{2^n-2}{n-1}\right)^{n-1} $$

$$n=2p $$ $$ \prod_{i=0}^{2p} {2p \choose i}=\prod_{i=1}^{2p-1} {2p \choose i}\leq {2p\choose p}^{2p-1} $$

$$ \left(\frac{4^p-2}{2p-1}\right)^{2p-1}=\left(\frac{1}{2p-1}\sum_{k=1}^{2p-1}2^k\right)^{2p-1} $$

It suffices to show that:

$$ {2p\choose p}\leq \frac{4^p-2}{2p-1} $$

$$ n=2p+1 $$

It suffices to show that:

$$ {2p+1\choose p}\leq \frac{4^p-1}{p}=\frac{1}{2p}\sum_{k=1}^{2n} 2^k $$


$$ \sqrt[n-1]{{n\choose 1}...{n\choose n-1}}\leq\frac{{n\choose 1}+...+{n\choose n-1}}{n-1}=\frac{2^n-2}{n-1} $$

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up vote 2 down vote accepted

Your approach will not work, because the central binomial coefficient satisfies $\binom{2p}{p} = \Theta \left( \frac{2^{2p}}{\sqrt{p}} \right)$, which is asymptotically larger than $\frac{2^{2p} - 2}{p-1}$.

HINT for the correct approach: Apply AM-GM inequality to the $n-1$ binomial coefficients $\binom{n}{i}$ where $1 \leqslant i \leqslant n-1$.

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Thanks! Actually the inequality is obvious, sorry... – Chon Dec 29 '11 at 14:19

Unless I made a mistake you won't be able to prove your result with this approach since your formula :

$$ {2p+1\choose p}\leq \frac{4^p-1}{p}=\frac{1}{2p}\sum_{k=1}^{2n} 2^k $$

is not true for p = 2

$$ {2*2+1\choose 2} = {5\choose 2} = 10 \geq \frac{4^2-1}{2} = 15/2 $$

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