Is there another way to write the product $\prod_{k=0}^n\left(k+\alpha\left(-1\right)^{k+1}\right)$? I have the following expression
$$ \prod_{k=0}^n \left(k + \alpha(-1)^{k+1}\right), $$
which is, for example, $(0-\alpha)(1+\alpha)(2-\alpha)$ for $n = 2$. Is there a way to write this using something like a factorial, instead of using Big Pi notation (or, similarly, a summation)? 
This is from a homework assignment; I'm sure the present form I've given is acceptable, but I'm curious if there are other nice forms.
 A: One can indeed rewrite your expression in terms of gamma function $\Gamma(z)$ or the so-called Pochhammer symbol
$$(z)_n=\frac{\Gamma(z+n)}{\Gamma(z)}=z(z+1)\ldots (z+n-1)=\prod_{k=0}^{n-1}(z+k).$$
Let us first consider the case of odd $n=2m-1$ with $m\in\mathbb{Z}_{>0}$. The product may then be rewritten as
\begin{align}
\prod_{k=0}^n\left(k+\alpha\left(-1\right)^{k+1}\right)&=\prod_{k=0}^{m-1}\left(2k-\alpha\right)\times \prod_{k=0}^{m-1}\left(2k+1+\alpha\right)=\tag{1}\\
&=2^{2m}\times\prod_{k=0}^{m-1}\left(-\frac{\alpha}{2}+k\right)\times
\prod_{k=0}^{m-1}\left(\frac{1+\alpha}{2}+k\right)=\\
&=2^{2m}\left(-\frac{\alpha}{2}\right)_m\left(\frac{1+\alpha}{2}\right)_m=\\
&=2^{2m}\frac{\Gamma\left(-\frac{\alpha}{2}+m\right)\Gamma\left(\frac{1+\alpha}{2}+m\right)}{\Gamma\left(-\frac{\alpha}{2}\right)\Gamma\left(\frac{1+\alpha}{2}\right)}.
\end{align}
For even $n$, the first product in (1) will contain one more factor, but the rest of manipulations is completely analogous. The results for both odd and even $n$ can be combined into a single formula
\begin{align}
\prod_{k=0}^n\left(k+\alpha\left(-1\right)^{k+1}\right)&=2^{n+1}\left(-\frac{\alpha}{2}\right)_{\lfloor{\frac{n}{2}+1}\rfloor}\left(\frac{1+\alpha}{2}\right)_{\lfloor\frac{n+1}{2}\rfloor}=\\
&=2^{n+1}\frac{\Gamma\left(-\frac{\alpha}{2}+\lfloor{\frac{n}{2}+1}\rfloor\right)\Gamma\left(\frac{1+\alpha}{2}+\lfloor\frac{n+1}{2}\rfloor\right)}{\Gamma\left(-\frac{\alpha}{2}\right)\Gamma\left(\frac{1+\alpha}{2}\right)}.
\end{align}
