Is there any formula for computing the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2k)$ I need to compute the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2n)$ in order to use the Stirling's formula to calculate, but I am stuck. Thanks!

I am a little bit confused now... should the last term of $w(n/2+1/2)$ be $(n+1-2n)$ or $(n+1-2k)$?
 A: Well, the comment thread here is long, and may have answers. I didn't check. 
For $n=2m$ even, and $k$ not so large that factors wander into negative territory, then you have:
$$\begin{align}
&\phantom{{}={}}(n+1)(n-1)\cdots(n+1-2k)\\
&=(2m+1)(2m-1)\cdots(2m+1-2k)\\
&=\frac{(2m+1)(2m)(2m-1)(2m-2)\cdots(2m+1-2k)(2m-2k)(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1}{\left[(2m)(2m-2)\cdots(2m-2k)\right]\left[(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1\right]}\\
&=\frac{(2m+1)!}{2^{k+1}\left[(m)(m-1)\cdots(m-k)\right]\left[(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1\right]}\\
&=\frac{(2m+1)!}{2^{k+1}\left[(m)(m-1)\cdots(m-k)\right](2m-2k-1)!}\\
&=\frac{(2m+1)!(m-k-1)!}{2^{k+1}m!(2m-2k-1)!}\\
&=\frac{(n+1)!\left(\frac{n}{2}-k-1\right)!}{2^{k+1}\left(\frac{n}{2}\right)!\left(n-2k-1\right)!}\\
\end{align}$$
A: While @alex.gordan beat me to a full proof that's simpler than mine functionally, I figured I would share the way I approached this problem! 
First note that $(a)_k = a(a+1)(a+2)\cdots(a+k-1)$. Thus, 
$$\begin{align}\left(\frac{1-n}{2}\right)_k  & = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+1\right)\cdots\left(\frac{1-n}{2}+k\right) \\
& = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+\frac{2}{2}\right)\left(\frac{1-n}{2}+\frac{4}{2}\right)\cdots\left(\frac{1-n}{2}+\frac{2k-2}{2}\right)\\
& = \left(\frac{1-n}{2}\right)\left(\frac{3-n}{2}\right)\left(\frac{5-n}{2}\right)\cdots\left(\frac{2k-1-n}{2}\right) \\& = \frac{(1-n)(3-n)(5-n)\cdots(2k-1-n)}{2^k}\end{align}$$
Let's now adjust this to fit your product. First let's remove the denominator
$$2^k\left(\frac{1-n}{2}\right)_k =(1-n)(3-n)(5-n)\cdots(2k-1-n)$$
Now let's pull out $k$ negative signs to flip each term
$$2^k(-1)^k\left(\frac{1-n}{2}\right)_k =(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
Now we simplify the powers and multiply by $(n+1)$
$$(-2)^k(n+1)\left(\frac{1-n}{2}\right)_k =(n+1)(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
We now use the definition $(a)_k = \frac{\Gamma(a+k)}{\Gamma(a)}$, let $n = 2m$, and transform this into
$$\frac{(-2)^k (n+1) \Gamma(\frac{2k+1-n}{2})}{\Gamma(\frac{1-n}{2})}$$
$$=(-2)^k (n+1)\frac{\Gamma(k-m+1/2)}{\Gamma(1/2-m)}$$
Now using the fact that $\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e}\right)^z$ (known as Stirling's approximation)
$$\sim (-2)^k (n+1)\frac{\sqrt{\frac{2\pi}{k-m+1/2}} \left(\frac{k-m+1/2}{e}\right)^{k-m+1/2}}{\sqrt{\frac{2\pi}{1/2-m}} \left(\frac{1/2-m}{e}\right)^{1/2-m}}$$
$$= \frac{(-2)^k (n+1)(k-m+1/2)}{e^k}\sqrt{\frac{1/2-m}{k-m+1/2}} \left(\frac{(k-m+1/2)}{(1/2-m)}\right)^{1/2-m}$$
A: Here is a closed form formula for the finite product 

$$ p=2^n(1-n)\frac{ \Gamma( \frac{(3+n) }{2}) } {\Gamma( \frac{(3-n) }{2}) }. $$

