Show that ${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n}$ Prove that 
$${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n},$$
$$\frac{1}n {2n -2 \choose n-1} 2^{-2n +1} = (-1)^{n-1} {\frac12 \choose n}.$$
The second part can be proved by replacing $n$ by $n-1$ in the first part. For the first part, I found that the right side is equal to ${n-1/2 \choose n}$, but when I expand the left side, I get something like
$$\frac{2n/4}{n}\frac{(2n-1)/4}{n-1}...\frac{(2n-n+1)/4}{1}$$
which does not look similar with ${n-1/2 \choose n}$.
I appreciate if you give some help. 
 A: First, you have
$$\begin{equation}\begin{aligned}
n! & = 1(2)(3)\ldots (n-1)(n) \\
& = \left(\frac{2}{2}\right)\left(\frac{4}{2}\right)\left(\frac{6}{2}\right)\ldots\left(\frac{2n-2}{2}\right)\left(\frac{2n}{2}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, since $n - \frac{1}{2} = \frac{2n - 1}{2}$, note you have
$$\begin{equation}\begin{aligned}
{\frac{2n - 1}{2} \choose n} & = \left(\frac{1}{n!}\right)\left(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n - 1}
{2}\right)\right) \\
& = \left(\frac{1}{(n!)^2}\right)\left(\frac{1}{2}\right)\left(\frac{2}{2}\right)\left(\frac{3}{2}\right)\left(\frac{4}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n - 1}
{2}\right)\left(\frac{2n}
{2}\right) \\
& = \left(\frac{1}{(n!)^2}\right)\frac{(2n)!}{2^{2n}} \\
& = 2^{-2n}\left(\frac{(2n)!}{(n!)^2}\right) \\
& = {2n \choose n} 2^{-2n}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
In the second line, I divided by the LHS of \eqref{eq1A} to get $\frac{1}{(n!)^2}$ and multiplied by the RHS of \eqref{eq1A} in the set of factors after that.
A: It might be easier to show
$$ 2^{2n}(-1)^n {-\frac12 \choose n} = {2n \choose n}$$
Indeed,
$$2^{2n}(-1)^n {-\frac12 \choose n} = 2^{2n}(-1)^n\frac{\prod_{i=0}^{n-1}\left(-\frac 12 - i\right)}{n!}= 2^n\frac{\prod_{i=0}^{n-1}\left(1+2i\right)}{n!}$$
$$= \frac{\prod_{i=1}^{n}2i\prod_{i=0}^{n-1}\left(1+2i\right)}{n!\cdot n!}=\frac{(2n)!}{n!\cdot n!}= {2n \choose n}$$
