The Problem
Show that
$$\binom{1/2}{k} = \frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}$$
My Work
$$\begin{align*}\frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k} &= \frac{(-1)^{k+1}(2k)!}{4^k(2k-1)(k!)^2}\\\\ &= \frac{(-1)^{k+1}(2k)(2k-1)(2k-2)\cdots(k+1)}{4^k(2k-1)(k!)}\\\\ &= \frac{(-1)^{k+1}(2k)(2k-2)\cdots(k+1)}{4^k(k!)}\\\\ &= \frac{(-1)(2k)(2k-2)\cdots(k+1)}{(k!)}\times \left(\frac{-1}{4}\right)^k \end{align*}$$
I basically don't know where to go from here. My denominator is almost right where I want it to be, my numerator is fairly out of order though. Any suggestions on what I should try next, or if I should take a different approach all together?
HINT: It’s very often a good idea to play with both sides, trying to get them to meet in the middle. In this case it’s a good idea to begin expanding the lefthand side:
$$\begin{align*} \binom{1/2}k&=\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\ldots\left(-\frac{2k-3}2\right)}{k!}\\\\ &=\frac{(-1)^{k-1}(2k-3)(2k-5)(2k-7)\ldots(3)(1)}{2^kk!}\;.\tag{1} \end{align*}$$
Now go back and take another look at the result of your first step in manipulating the righthand side:
$$\frac{(-1)^{k+1}(2k)!}{4^k(2k-1)(k!)^2}\;.\tag{2}$$
Clearly $(-1)^{k+1}=(-1)^{k-1}$. The denominator of $(1)$ suggests splitting the denominator of $(2)$ as $(2^kk!)\cdot\big((2k-1)2^kk!\big)$ and trying to cancel the $(2k-1)2^kk!$ with something in the numerator. Thus, it looks as if you want to try to prove that
$$(2k-3)(2k-5)(2k-7)\ldots(3)(1)=\frac{(2k)!}{(2k-1)2^kk!}\;.$$
This isn’t too hard; try to combine $2^kk!$ in the denominator into a single product of $k$ integers.