Question 4
The product $$
\begin{gathered} \prod_{k=4}^{63} \frac{\log_k (5^{k^2 - 1})}{\log_{k+1} (5^{k^2 - 4})} \\ = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \\ \quad {}\cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})} \\ \quad {}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \\ \quad \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})} \end{gathered}
$$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
\begin{gathered} \prod_{k=4}^{63} \frac{\log_k (5^{k^2 - 1})}{\log_{k+1} (5^{k^2 - 4})} \\ = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \\ \quad {}\cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})} \\ \quad {}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \\ \quad \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})} \end{gathered}
$$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$