Wallis Product (Long infinite Product) I'm almost finished proving Wallis product in a given question. But the last step is:
What am I doing wrong? Am I on the right track? Hints?
Show that $$\prod_{k=1}^{m} \left(\frac{(2k-1)(2k+1)}{(2k)^2}\right) =  \frac{(2m+1)((2m!))^2}{2^{4m}(m!)^4}$$
Currently I have tried expanding a few terms as so:
$$\frac{1 * 3}{2^2} * \frac{3 * 5}{4^2} * \frac{5 * 7}{6^2}*.........*\frac{(2k-7)(2k-5)}{(2(k-3))^2} *\frac{(2k-5)(2k-3)}{(2(k-2))^2}*\frac{(2k-3)(2k-1)}{(2(k-1))^2}*\frac{(2k-1)(2k+1)}{(2k)^2}$$
I note that the denominator is $$4*m = 2^{2m}$$
For the numerator I get $(2m+1)*[(2m-1)!]^2$
Which is not really close to what I want...
I want to try this manipulation
$$\frac{1 * 3}{2^2} *\frac{3^2}{3^2} * \frac{3 * 5}{4^2} * \frac{5^2}{5^2} \frac{5 * 7}{6^2}*.........*\frac{(2k-7)(2k-5)}{(2(k-3))^2} *\frac{(2k-5)^2}{(2k-5)^2}*\frac{(2k-5)(2k-3)}{(2(k-2))^2}*\frac{(2k-3)^2}{(2k-3)^2}*\frac{(2k-3)(2k-1)}{(2(k-1))^2}*\frac{(2k-1)^2}{(2k-1)^2}*\frac{(2k-1)(2k+1)}{(2k)^2}$$
Note: Excuse my errors: It's 3 AM here, I am brain dead. 

You are quite close. If we multiply out, then on top we get $2m+1$ times the product of the squares of the odd numbers up to $2m-1$. So we get
$$(2m+1) \left(1\cdot 3\cdot 5\cdots(2m-1)\right)^2.$$
"Improve" $1\cdot 3\cdot 5\cdot (2m-1)$ to $(2m)!$  by filling in the even numbers, that is, by multiplying by $2\cdot 4\cdot 6\cdots (2m)$.  We must also divide by this number, which is $2^m m!$. With this manipulation, the original top becomes 
$$(2m+1)\frac{((2m)!)^2}{2^{2m}(m!)^2}.\tag{1}$$ 
The manipulation of the original bottom is simpler: the product $\prod (2k)^2$ is just $$2^{2k}(m!)^2.\tag{2}$$
Finally, we divide (1) by (2). 
