Use the Euler product formula

$$\Gamma(z)=\lim _{n \rightarrow \infty} \frac{n ! n^{z}}{z(z+1) \ldots(z+n)}$$

to show that:

(i) $\Gamma(z+1)=z \Gamma(z)$;

(ii) $\frac{1}{\Gamma(z)}=z e^{\gamma z} \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right) e^{-z / k}$, where $\gamma=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n\right)$.

Deduce that

$$\frac{d}{d z} \log (\Gamma(z))=-\gamma-\frac{1}{z}+z \sum_{k=1}^{\infty} \frac{1}{k(z+k)}$$