Proving $ \phi(mn)=\phi(m)\phi(n) \frac k{\phi(k)}$

Note that $$\frac{\phi{(m)}}{m}=\prod_{p|m}(1-\frac{1}{p})$$

Thus

\begin{align} \frac{\phi{(mn)}}{mn}=\prod_{p|mn}(1-\frac{1}{p}) & =\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|m, p|n}(1-\frac{1}{p})} \\ &=\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|k}(1-\frac{1}{p})} \\ & =\frac{\phi{(m)}}{m}\frac{\phi{(n)}}{n}\frac{k}{\phi{(k)}} \end{align}

Multiplying by $mn$ gives the desired equality.

If $m$ and $n$ are coprime, then it is easy to show that $\phi(mn)=\phi(m)\phi(n)$. So we assume that $m=p^a$and $n=p^b$. Then $\phi(mn)=\phi(p^{a+b})=p^{a+b-1}(p-1)=p^{a-1}(p-1)p^{b-1}(p-1)\frac{p}{p-1}$. So the result follows.

P.S. When $m$ and $n$ are relatively prime, if $k$ is coprime to $mn$ then $k$ is prime to $m$ and $n$. Conversely, if $k_1$ is prime to $m$ and $k_2$ is prime to $n$, then there is a number prime to $mn$ and $\equiv k_1\pmod m$ and $\equiv k_2\pmod n$. And hence $\phi(mn)=\phi(m)\phi(n)$.

Point to me any error that occurs, thanks.