Answer
$1$
Work Step by Step
Using The Change-of-Base Formula ($\log_a{b}=\frac{\log_c{b}}{\log_c{a}}$):$\log_23\log_34...\log_n{n+1}\log_{n+1}2=\frac{\log3}{\log2}\frac{\log4}{\log3}...\frac{\log n+1}{\log n}\frac{\log2}{\log n+1}$.
We can see that the terms cancel, thus the product is $1$.
