To show: $$\forall k\in\mathbb{N}_0:\binom{2k}{k}=(-4)^k\binom{-0.5}{k}$$
$\underline{Basecase}$
$k=0$ $$\binom{2\cdot 0}{0}=\frac{0!}{0!0!}=1=\frac{(-0.5)!}{(-0.5)!}=1\cdot \frac{(-0.5)!}{0!(-0.5-0)!}=(-4)^0\cdot \binom{-0.5}{0}$$${\huge \checkmark}$
$\underline{Assumption}$
For some $k\in\mathbb{N}_0$ the following holds: $$\binom{2k}{k}=(-4)^k\binom{-0.5}{k}$$ Which is equivalent to $$\frac{(2k)!}{k!k!}=(-4)^k\cdot \frac{(-0.5)!}{k!(-0.5-k)!}$$ $\underline{Claim}$
For $(k+1)\in\mathbb{N}_0$ the following holds: $$\binom{2(k+1)}{k+1}=(-4)^{k+1}\binom{-0.5}{k+1}$$ $\underline{Step}$
We show that the equality $$\binom{2(k+1)}{k+1}=(-4)^{k+1}\binom{-0.5}{k+1}$$holds.
First we expand it like the assumption: $$\frac{(2k+2)!}{(k+1)!(k+1)!}=(-4)^{k+1}\cdot \frac{(-0.5)!}{(k+1)!(-0.5-k-1)!}$$ $$\iff \frac{(2k)!}{(k)!(k)!}\cdot\frac{(2k+1)(2k+2)}{(k+1)^2}=(-4)^{k}\cdot \frac{(-0.5)!}{(k)!(-0.5-k)!}\cdot (-4) \frac{(-0.5-k)}{(k+1)}$$ Now we use our assumption $$\iff \frac{(2k+1)(2k+2)}{(k+1)^2}=(-4) \frac{(-0.5-k)}{(k+1)}$$ $$\iff (2k+1)(2k+2)=(4k+2)(k+1)$$ $$\iff 4k^2+4k+2k+2=4k^2+4k+2k+2$$
Is the following right? $$(-0.5-k-1)!=\frac{(-0.5-k)!}{(-0.5-k)}$$
We typically use the gamma function to extend the factorials to all of $\mathbb{C}$; thus including the negatives, as you desire. However, the gamma function is shifted down by 1; that is,
$$\Gamma(k) = (k-1)!$$
for positive integers $k$. The purpose of this function is to satisfy the properties of the factorial, while being a smooth curve. In fact, one of the restrictions that was applied to find the gamma function was the restriction
$$\Gamma(k+1) = k\Gamma(k)$$
Thus, given $k! = \Gamma(k+1)$, we find that
$$k! = \Gamma(k+1) = (k)\Gamma(k) = k(k-1)!$$
and as such,
$$(k-1)! = \frac{k!}{k}$$
as requested. Note that this holds only for $k \notin \mathbb{Z}_-$.