I would like to prove the following: given the sequence $$a\_m = \begin{cases}1 & \text{if $m = 0, 1$} \\\frac{(\alpha+2)(\alpha+4)\cdots(\alpha+2(m-1))}{(\alpha+1)(\alpha+2)\cdots(\alpha+m-1)} & \text{otherwise}\end{cases}$$ then, $$\sum\_{m=0}^{n}(-1)^m\binom{n}{m}a\_m = \begin{cases}\frac{(n-1)!!}{(\alpha+1)(\alpha+3)\cdots(\alpha+n-1)}& \text{if $n$ is even and $\ge 2$} \\0 & \text{if $n$ is odd}\end{cases}$$

I understand that the left hand side in the last equation is the $n$-th difference $\Delta^na\_0$, so basically I tried to compute each difference but it seems that a general formula is hard to find for $\Delta^na\_k$. Can you help me? I have seen often that such problems can be attacked with special functions but unfortunately I don't have any experience :(. What do you suggest? Any hint would be greatly appreciated :)

EDIT: So here is the original problem: show that $$e^{-x}\left[1+x+\frac{\alpha+2}{\alpha+1}\cdot\frac{x^2}{2!}+\frac{(\alpha+2)(\alpha+4)}{(\alpha+1)(\alpha+2)}\cdot\frac{x^3}{3!}+\ldots\right]=$$$$1+\frac1{\alpha+1}\frac{x^2}{2^1\cdot1!}+\frac1{(\alpha+1)(\alpha+3)}\frac{x^4}{2^2\cdot2!}+\ldots$$. Doing the product one just compares the two series and obtains my equation above (if I interpreted correctly the pattern of the two series).