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:
$$ {2p+1\choose p}\leq \frac{4^p-1}{p}=\frac{1}{2p}\sum_{k=1}^{2n} 2^k $$
AM-GM:
$$ \sqrt[n-1]{{n\choose 1}...{n\choose n-1}}\leq\frac{{n\choose 1}+...+{n\choose n-1}}{n-1}=\frac{2^n-2}{n-1} $$