Prove that $f^{2016}(z)=z(z-1)\ldots(z-2015)$ for an analytic $f$.

Question

Let $\Omega= \mathbb{C}\setminus \{x+iy:~x\in \mathbb{N},~y\geq 0\}$. Prove there there is an analytic function $f$ on $\Omega$, such that

$$f^{2016}(z)=z(z-1)\ldots(z-2015).$$

I don't know where to start.

Thanks a lot in advance!

Answer 1

Hint: $\Omega$ is simply connected and $z(z-1)\ldots(z-2015)$ has no zero in $\Omega$