7.257. $\log _{2} 3 \cdot \log _{3} 4 \cdot \log _{4} 5 \cdot \ldots \cdot \log _{n}(n+1)=10(n \in N)$.
Solution.

Let's switch to base 2.

$\log _{2} 3 \cdot \frac{\log _{2} 4}{\log _{2} 3} \cdot \frac{\log _{2} 5}{\log _{2} 4} \cdot \ldots \cdot \frac{\log _{2}(n+1)}{\log _{2} n}=10 \Leftrightarrow \log _{2}(n+1)=10 \Leftrightarrow$

$\Leftrightarrow n+1=2^{10} \Leftrightarrow n=1024-1=1023$.

Answer: 1023.