It is well-known that the Gamma function $\Gamma$ satisfies $$\Gamma(x+1)=x\Gamma(x)\tag{1}$$ for all $x\in \mathbb{R}\setminus \{0,-1,-2,\dots\}$. I have shown that $\log\Gamma$ is differentiable on $(0,\infty)$. Since $\Gamma = \exp(\log \Gamma)$, we have that $\Gamma$ is differentiable on $(0,\infty)$. Now, the question is:
Question: How to conclude from $(1)$ that $\Gamma$ is differentiable on whole $\mathbb{R}\setminus \{0,-1,-2,\dots\}$?
It may seem too obvious, but how would you explain it mathematically?