Proving a complex binomial identity

We apply the Cauchy product formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^n](1+z)^\alpha=\binom{\alpha}{n} \end{align*}

We obtain for $\alpha\in\mathbb{C}$ and $n\in\mathbb{N}$: \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k(k+1)\binom{\alpha+2}{n-k}}\\ &=\sum_{k=0}^n\left([z^k]\frac{1}{(1+z)^2}\right)\left([z^{n-k}](1+z)^{\alpha+2}\right)\tag{1}\\ &=[z^n](1+z)^{\alpha}\\ &\,\,\color{blue}{=\binom{\alpha}{n}} \end{align*} and the claim follows.

Comment:

• In (1) we use the Cauchy product formula\begin{align*} A(z)&=\sum_{k=0}^\infty a_kz^k,\qquad B(z)=\sum_{j=0}^\infty b_jz^j\\ A(t)B(t)&=\sum_{n=0}^\infty\left(\sum_{{k+j=n}\atop{k,j\geq 0}}a_kb_j\right)t^n=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k b_{n-k}\right)t^n\\ &=\sum_{n=0}^\infty\sum_{k=0}^n \left([z^k]A(z)\right)\left([z^{n-k}]B(z)\right)t^n \end{align*}

  In the case above we have \begin{align*} A(z)&=\sum_{k=0}^\infty (-1)^k(k+1)z^k=\frac{1}{(1+z)^2}\\ B(z)&=\sum_{k=0}^\infty \binom{\alpha+2}{k}z^k=(1+z)^{\alpha+2} \end{align*}