show that $e=(\frac{2}{1})^{\frac{1}{1}}(\frac{4}{3})^{\frac{1}{2}}(\frac{6\cdot8}{5\cdot7})^{\frac{1}{4}}...$ I found 
$$e=(\frac{2}{1})^{\frac{1}{1}}(\frac{4}{3})^{\frac{1}{2}}(\frac{6\cdot8}{5\cdot7})^{\frac{1}{4}}(\frac{10\cdot12\cdot14\cdot16}{9\cdot11\cdot13\cdot15})^{\frac{1}{8}}.....$$
easily you can find the relation between the number:
$$a_n=\left ( \frac{\frac{(2^n)!!}{(2^{n-1})!!}}{\frac{(2^n-1)!!}{(2^{n-1}-1)!!}} \right )^{2^{-n+1}}:n>0$$
then using $$(2k)!!=2^k(k)!$$
$$(2k-1)!!=\frac{(2k)!}{2^k(k)!}$$
hence 
$$\Rightarrow a_n=\left ( \frac{2^{2^{n-1}}(2^{n-1}!)^3}{(2^{n-2}!)^2(2^{n}!)} \right )^{2^{-n+1}}$$
so $$e\stackrel{?}{=}\prod_{n=1}^{\infty}\left ( \frac{2^{2^{n-1}}(2^{n-1}!)^3}{(2^{n-2}!)^2(2^{n}!)} \right )^{2^{-n+1}}$$ 
and I think to take log of both side then try to prove that R.H.S $=1$

You have done the hard part, the rest is easy:
$$\log a_n=\frac{1}{2^{n-1}}\log \left ( \frac{2^{2^{n-1}}(2^{n-1}!)^3}{(2^{n-2}!)^2(2^{n}!)} \right ) \\= \log 2 +\frac{1}{2^{n-1}}\left(3\log (2^{n-1}!) - 2\log (2^{n-2}!) - 2\log (2^{n}!)\right) \\ = \log 2 + \frac{1}{2^{n-2}}\left(\log (2^{n-1}!) - \log (2^{n-2}!)\right) - \frac{1}{2^{n-1}}\left(\log (2^{n}!) - \log (2^{n-1}!)\right)$$
Thus, except for the $\log 2$ part the rest of the sum telescopes:
$$\sum\limits_{n=1}^{N} \log a_n = (N+1)\log 2 - \frac{1}{2^{N-1}}\left(\log (2^{N}!) - \log (2^{N-1}!)\right) \\= 2\log 2 -\frac{1}{x_N}\log \frac{(2x_N)!}{x_N^{x_N}x_N!}$$
Where, $x_N = 2^{N-1}$.
Now, $$\lim\limits_{n \to \infty} \sqrt[n]{\frac{(2n)!}{n^nn!}} = \lim\limits_{n \to \infty}\frac{(2n+2)!n^nn!}{(n+1)^{n+1}(n+1)!(2n!)} = \log \frac{4}{e}$$
Which implies $$\sum\limits_{n=1}^{\infty} \log a_n = 1$$
