Problem 473

Since $\sqrt{\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \cdots \times \frac{n-1}{n}} = \frac{1}{8}$, then squaring both sides, we obtain $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \cdots \times \frac{n-1}{n} = \frac{1}{64}$. Simplifying the left side, we obtain $\frac{1 \times 2 \times 3 \times 4 \times \cdots \times (n-1)}{2 \times 3 \times 4 \times 5 \times \cdots \times n} = \frac{1}{64}$ or $\frac{1 \times (2 \times 3 \times 4 \times \cdots \times (n-1))}{(2 \times 3 \times 4 \times \cdots \times (n-1)) \times n} = \frac{1}{64}$ and so $\frac{1}{n} = \frac{1}{64}$ which means that $n = 64$.