Proving a complex binomial identity I would like to prove an identity:
$$\binom{\alpha}{n} = \sum_{k=0}^n(-1)^k(k+1)\binom{\alpha + 2}{n-k}$$
Where $\alpha$ is complex.
I have already found that if you have two sequences related by the identity $$b_n = \sum_{k=0}^n(-1)^{k}(1+k)a_{n-k}$$ you can write the generating function for $b_n$ (which I'll write as $B(z)$) in terms of the generating function for $a_n$ as follows:
$$B(z) = \frac{A(z)}{(1+z)^2}$$
How do I prove this identity, using the above fact?
Thanks in advance!

The classic recursion for Pascal's triangle is
 $\;{\alpha\choose n}+{\alpha\choose n-1}={\alpha+1\choose n},\;$ applied twice giving $\;{\alpha+2\choose n}.\;$ Also
 $\;(1+z)B(z)=A(z)\;$ is equivalent to $\;a_n=b_n+b_{n-1},\;$ and applied twice  we are done.
