smc

The product $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdots \left(1-\frac{1}{n}\right)$ can be simplified to $\frac{2\ast 3\ast 4\cdots \ast (n-2)\ast (n-1)}{3\ast 4\ast 5\cdots \ast (n-1)\ast n}$, which, through telescoping, ultimately works out to $\boxed{\frac{2}{n}}$.