Adieu 2017

Note that $\int_0^\infty \ln(1 - e^{-my})\,dy \; = \; -\sum_{n \ge 1} \frac{1}{n}\int_0^\infty e^{-mny}\,dy \; = \; -\sum_{n \ge 1} \frac{1}{mn^2} \; = \; -\frac{\zeta(2)}{m}$ for any $m > 0$ , and hence $\begin{aligned} A_n & = \; \int_{-\infty}^0 \ln(1 + e^x + e^{2x} + \cdots + e^{nx})\,dx \; = \; \int_{-\infty}^0 \ln\left(\frac{e^{(n+1)x}-1}{e^x-1}\right)\,dx \\ & = \; \int_0^\infty \ln\left(\frac{1 - e^{-(n+1)y}}{1 - e^{-y}}\right)\,dy \; = \; \int_0^\infty \ln(1 - e^{-(n+1)y})\,dy - \int_0^\infty \ln(1 - e^{-y})\,dy \\ & = \; \zeta(2) - \frac{\zeta(2)}{n+1} \; = \; \frac{n}{n+1}\zeta(2) \end{aligned}$ and hence $\prod_{n=1}^{2017} A_n \; = \; \frac{\zeta(2)^{2017}}{2018}$ Using the Euler identity $\prod_p \left(1 - p^{-2}\right) \;= \; \zeta(2)^{-1}$ (and assuming that the numerator in the question is asking for the product over all primes), we deduce that the answer is $\boxed{2018}$ .