Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to 
$$\Gamma\left ( -k+\frac{1}{2} \right )=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}.\tag{$k\geq 1$}$$
If possible, I'd like you to give a hint. If it is not possible to use Legendre's duplication formula, then I tried this way, for $n\geq 1$,
\begin{align*}
\Gamma\left ( -n+\frac{1}{2} \right )&=\left ( -n-\frac{1}{2} \right )\Gamma\left ( -n-\frac{1}{2} \right )\\
&=\dots\\
&=\left ( \frac{1}{2} \right )\left ( \frac{3}{2} \right )\cdots\left ( \frac{-2n-3}{2} \right )\left ( \frac{-2n-1}{2} \right )\Gamma\left ( \frac{1}{2} \right )
\end{align*}
and not anymore. How do I count how many factors there are to right side beside $\Gamma(1/2)$?

Is this Possible?
Since, 
$\Gamma(a)=\frac{\Gamma(a+n)}{(a)_n}$ You can deduce that
\begin{eqnarray*}
 \Gamma(-m+\frac{1}{2})&=&\frac{\Gamma(\frac{1}{2})}{(-m+\frac{1}{2})_m}\\
 &=&\frac{\sqrt{\pi}}{\left(\frac{-2m+1}{2}\right)\cdot\left(\frac{-2m+3}{2}\right)\cdot\left(\frac{-2m+5}{2}\right)\cdots\left(\frac{1}{2}\right)}\\
 &=&\frac{(-1)^m\cdot 2^m\cdot\sqrt{\pi}}{(2m-1)\cdot(2m-3)\cdots 3\cdot 2\cdot 1}
\end{eqnarray*}
