Proof Given that
$$\begin{aligned} f(3) f(7) & =f(21)f(2)=2, \text{ hence } f(3)=3. \end{aligned}$$
If the proposition is not true, let the smallest positive integer for which $f(n) \neq n$ be $n_{0} \geqslant 4$. Since $f\left(n_{0}\right)>f\left(n_{0}-1\right)=n_{0}-1$, it follows that $f\left(n_{0}\right)>n_{0}$. Furthermore, because $f(n)$ is strictly increasing, we have
$$f(n)>n \text{, when } n \geqslant n_{0} \text{. }$$

When $n_{0}$ is odd, 2 and $n_{0}-2$ are coprime, so we have
$$\begin{aligned} f\left(2\left(n_{0}-2\right)\right) & =f(2) f\left(n_{0}-2\right) \\ & =2\left(n_{0}-2\right) . \end{aligned}$$

Since $n_{0} \geqslant 4$, it follows that $2\left(n_{0}-2\right) \geqslant n_{0}$, which contradicts (1). When $n_{0}$ is even, 2 and $n_{0}-1$ are coprime, so we have
$$\begin{aligned} f\left(2\left(n_{0}-1\right)\right) & =f(2) f\left(n_{0}-1\right) \\ & =2\left(n_{0}-1\right) . \end{aligned}$$

Clearly, $2\left(n_{0}-1\right)>n_{0}$, which contradicts (1).