Wallis Product (Long infinite Product)

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.