For which natural numbers $n$ is it true that

$$
\log _{2} 3 \cdot \log _{3} 4 \cdot \log _{4} 5 \cdot \ldots \cdot \log _{n}(n+1)=10 ?
$$
We rewrite the factors on the left side of the equality in base-10 logarithms:

$$
\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdot \frac{\lg 5}{\lg 4} \cdot \ldots \cdot \frac{\lg n}{\lg (n-1)} \cdot \frac{\lg (n+1)}{\lg n}=10
$$

After performing the simplifications, we get that \(\frac{\lg (n+1)}{\lg 2}=10\), rearranging gives \(\lg (n+1)=\lg 2^{10}\), hence \(n=2^{10}-1=1023\).