I was reading proof of Binet's first expression for $\log \Gamma(z)$; that is for $\Re z > 0$, $$ \log \Gamma(z) = \left(z - \frac 1 2\right)\log z - z + \frac 1 2 \log (2\pi) + \int_0^\infty \left(\frac 1 2 - \frac 1 t + \frac 1 {e^t - 1} \frac {e^{-tz}}t \right) \operatorname{d}\!t. $$
One of the steps was using the following: for $\Re z > 0$, $$\log\Gamma(z+1) = \log z + \log \Gamma(z). \qquad\qquad(*)$$ At first glance it seems obvious, since $\Gamma(z+1) = z\Gamma(z)$. But equality $\log(uv) = \log u + \log v$ holds only up to difference of $2\pi i$ in general. Here $\log$ is taken to be discontinuous at non positive real numbers.
If $\Re \Gamma(z) > 0$, then there is no additional $\pm 2\pi i$, but this inequality is not always true in our domain.
I was told that Weierstrass form could be helpful, but I don't see how. How to show $(*)$?