Suppose we defined the Gamma function $$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac zn\right)e^{-z/n}$$ where $\gamma$ is just a constant. I want to prove that $\Gamma(1)=1$, so I need to prove that $$e^{-\gamma}=\prod_{n=1}^\infty\left(1+\frac1n\right)e^{-1/n}\tag{*}$$
From definition, by taking logarithm and then differentiating, we arrive to $$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{n=1}^\infty\left(\frac1{z+n}-\frac1n\right)\tag{1}$$ which implies that $$\frac{\Gamma'(z+1)}{\Gamma(z+1)}-\frac{\Gamma'(z)}{\Gamma(z)}-\frac1z=0\tag{2}$$ taking integral we arrive to the following identity $$\Gamma(z+1)=Cz\Gamma(z)\tag{3}$$ where $C$ is constant and since $\lim_{z\to0}z\Gamma(z)=1$, so $$C=\Gamma(1)\tag{4}$$ I tried to from identities $(1-4)$ arrive to $(*)$, but failed, can anyone help me, please?