I want to show the following infinite product, $$\frac{\psi(z)}{\Gamma(z)}=-e^{2\gamma z}\prod_{k=0}^{\infty}\left(1-\frac{z}{x_k}\right)e^{z/x_k}$$ where $x_k$ is the $k$th root of the digamma function.
This is what I've tried. As noted on the wiki, $$\frac{\psi(z)}{\Gamma(z)}=\frac{\Gamma'(z)}{\Gamma(z)^2}=-\frac{d}{dz}\frac{1}{\Gamma(z)}.$$ By Weierstrass's infinite product expansion of the reciprocal gamma, $$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n}$$ taking the derivative, $$\begin{align*} \frac{d}{dz}\frac{1}{\Gamma(z)}&=\frac{d}{dz}\left[ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-z/k}\right]\\&=(1+\gamma z)e^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-z/k}-ze^{\gamma z}\sum_{k=1}^\infty\left(\frac{e^{-z/k}}{k^2}\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-z/n}\right) \\ &=\frac{(1+\gamma z)}{z\Gamma(z)}-ze^{\gamma z}\sum_{k=1}^\infty\left(\frac{e^{-z/k}}{k^2}\frac{1}{ze^{\gamma z}\Gamma(z)}\right) \\ &=\frac{(1+\gamma z)}{z\Gamma(z)}-\frac{1}{\Gamma(z)}\sum_{k=1}^\infty\left(\frac{e^{-z/k}}{k^2}\right) \end{align*}$$ and I am stuck. More precisely, I do not know how to make the connection to the roots of $\psi$.