I have a (newbie) doubt about the value of the regularized product of $p-1$ (being $p$ a prime number):
Is it the same $$\prod^\infty{(p^{-s}-1)}$$ as $$\prod^\infty{(1-p^{-s})}$$ ? Why/why not?
On the one hand, don't think so, as, for example, taking $s=-1$ on the first equation we would get $$\prod^\infty{(p-1)}=\frac{4\pi^2}{\zeta{(1)}}=0$$ but taking it on the second equation we would get $$\prod^\infty{1-p^{-s}}=({\prod^\infty{(\frac{1}{1-p})}})^{-1} = {\zeta{(-1)}}^{-1}=-12$$ On the other hand, it seems obvious for me that $|1-a|=|a-1|$