Ln of gamma function

Question

Im having trouble with one step of the calculation of the derivative of the $\ln(\Gamma(z))$: $$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod(1+\frac{z}{k})e^{-z/k},$$ then taking the logarithm of both sides I arrived at:$$\ln(\Gamma(z))=-\ln(z) - \gamma z - \sum \{ \ln(1+\frac{z}{k}) - \frac{z}{k} \} $$ but the correct expression is $$\ln(\Gamma(z))=-\ln(z) - \gamma z - \sum \ln\{ (1+\frac{z}{k}) - \frac{z}{k} \}$$ How works that step???