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 $\Gamma(1/2)$?

I think that the reflection formula
$$
\Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin{(\pi z)}} \qquad z \not \in \mathbb{Z},
$$
should work better here, given that you already have computed $\displaystyle{\Gamma\left(k + \frac{1}{2}  \right)}$.
