EDIT notes: Deleted irrelevant information
I'm asked to prove that $z\Delta(z+1)=\Delta(z)$, where $\Delta(z)\equiv ze^{\gamma z}\prod\limits_{m=1}^\infty(1+z/m)e^{-z/m}$ and $\gamma =\lim_{n\to\infty} 1+ 1/2 + 1/3 + ... + 1/n-\log n$.
My idea was to look at finite n both in $\gamma$ and in the product. First I assumed the theorem to be true, thus getting $\Delta(z+1)=\Delta(z)/z$. I divided both sides by $e^{\gamma z}$ and then took the logarithm. From here, managed to prove equivalence. Is there another, "cleaner" proof?
Consider the telescoping sum $$\sum_{m=1}^n \left(\log\left(1+\frac{z}{m+1}\right)-\log\left(1+\frac{z}{m}\right)\right) = \log\left(1+\frac{z}{n+1}\right) - \log(1+z)$$ Then $$- \log(1+z) = \sum_{m=1}^\infty \left(\log\left(1+\frac{z}{m+1}\right)-\log\left(1+\frac{z}{m}\right)\right) = \sum_{m=1}^\infty \left(\log\left(\frac{z+m+1}{z+m}\right)-\log\left(\frac{m+1}{m}\right)\right) = \sum_{m=1}^\infty \left(\log\left(\frac{z+m+1}{z+m}\right)-\frac{1}{m}+\frac{1}{m}-\log\left(\frac{m+1}{m}\right)\right) = \sum_{m=1}^\infty \left(\log\left(\frac{1+(z+1)/m}{1+z/m}\right)-\frac{1}{m}\right)+\gamma$$ Apply exponential to get $$\frac{1}{1+z} = e^\gamma \prod_{m=1}^\infty\frac{1+(z+1)/m}{1+z/m}e^{-1/m}=\frac{e^{\gamma(z+1)}}{e^{\gamma z}}\prod_{m=1}^\infty \frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$ Then $$z e^{\gamma z} \prod_{m=1}^\infty (1+z/m)e^{-z/m} = z(z+1) e^{\gamma(z+1)}\prod_{m=1}^\infty (1+(z+1)/m)e^{-(z+1)/m}$$ and we are done