An identity for $\binom{-1/2}{n}$

More than a hint: Recall that the falling factorial is defined as $x^\underline{k}=\frac{x!}{(x-k)!}=x(x-1)\cdots (x-k+1),$ and the rising factorial is defined as $x^\overline{k}=\frac{(x+k-1)!}{(x-1)!}=x(x+1)\cdots (x+k-1).$

This is the binomial theorem $$(x+y)^n=\sum _{k=0}^n\binom{n}{k}x^ky^{n-k}.$$ The beautiful part is that this keeps happening for $$(x+y)^{\underline{n}}=\sum _{k=0}^n\binom{n}{k}x^{\underline{k}}y^{\underline{n-k}},$$ notice that if you take $x=n,y=-1/2-n,$ then $$(-1/2)^{\underline{n}}=\sum _{p=0}^n\binom{n}{p}n^{\underline{n-p}}(-1/2-n)^{\underline{p}}.$$ Then you have to check that $(-x)^{\underline{k}}=(-1)^k(x)^{\overline{k}}.$

Let $ \alpha $, $ \beta \in\mathbb{R} $. we know that, for any $ x\in\mathcal{B}\left(0,1\right)=\left(-1,1\right) $, we have : $$ \left(1+x\right)^{\alpha}=\sum_{n=0}^{+\infty}{\binom{\alpha}{n}x^{n}} $$ $$ \left(1+x\right)^{\beta}=\sum_{n=0}^{+\infty}{\binom{\beta}{n}x^{n}} $$

Which means : $$ \small\sum_{n=0}^{+\infty}{\binom{\alpha +\beta}{n}x^{n}}=\left(1+x\right)^{\alpha + \beta}=\left(\sum_{n=0}^{+\infty}{\binom{\alpha}{n}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\binom{\beta}{n}x^{n}}\right)=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{\alpha}{k}\binom{\beta}{n-k}}\right)x^{n}} $$

Thus, for any $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{\alpha}{k}\binom{\beta}{n-k}}=\binom{\alpha +\beta}{n} $$

Setting $ \beta = n $, $ \alpha = -n-\frac{1}{2} $, we get : $$ \sum_{k=0}^{n}{\binom{-n-\frac{1}{2}}{k}\binom{n}{n-k}}=\binom{-\frac{1}{2}}{n} $$

By definition $ \binom{-n-\frac{1}{2}}{k}=\frac{1}{k!}\prod\limits_{j=0}^{k-1}{\left(-n-\frac{1}{2}-j\right)} $, we also have that : $ \binom{n}{k}=\binom{n}{n-k} $. Thus : $$ \sum_{k=0}^{n}{\binom{n}{k}\frac{1}{k!}\prod_{j=0}^{k-1}{\left(-n-\frac{1}{2}-j\right)}}=\binom{-\frac{1}{2}}{n} =\frac{\left(-1\right)^{n}}{4^{n}}\binom{2n}{n}$$