Welcome 2016! Part 5

$\large \displaystyle \prod_{n=2}^{1023} \log_{n}(n+1)$

$= \large \log_{2}3\times \log_{3}4 \times \ldots \times \log_{1023}1024$ Using logarithm property , $\log_b a = \dfrac{\ln a}{\ln b}$
$= \large \dfrac{\ln 3}{\ln 2} \times \dfrac{\ln 4}{\ln 3} \times \ldots \times \dfrac{\ln 1024}{\ln 1023}$ $= \large \dfrac{\ln 1024}{\ln 2}= \log_2 1024= 10$

$\large \displaystyle \prod_{n=2}^{1023} \log_{n}(n+1)$

= $\large (\log_{2}3)(\log_{3}4) \ldots (\log_{1023}1024$ )

By the chain rule of logarithms $(\large \log_{a}b \times \log_{b}c = \log_{a}c)$

= $\large \log_{2}1024 = 10$

write as $\prod_{n=2}^{1023} \left(\dfrac{\ln(n+1)}{\ln(n)}\right)$ this easily telescopes to $\dfrac{1}{\ln(2)}\ln(1023+1)=\log_2 (1024)=10$