# Proving multiplicitive properties of Euler's phi function

I'm currently working on this problem:

We know that if $m,n \in \mathbb{N}$ are pair wise, relatively prime integers, then $\phi(m,n) = \phi(m) \times \phi(n)$ Prove by induction that if $m_1,...,m_r$ are pair-wise relatively prime positive integers, then $\phi(m_1\cdot m_r) = \phi(m_1)\cdot\cdot\cdot\phi(m_r).$

What I have so far:

For our base case, let $r=2$

Then $\phi(m_1\cdot m_r)=\phi(m_1)\cdot\cdot\cdot\phi(m_r)$ is the same as $\phi(m_1\cdot m_2)=\phi(m_1)\phi(m_2)$ which is the given property, so we know that the base case holds.

The inductive step is what I'm not sure about. Any pointers regarding how I could proceed?

We know that $\varphi(n)=\left|\mathbb{Z}/(n\mathbb{Z})^*\right|$.
Assuming that $m,n$ are coprime, by the Chinese remainder theorem we have $$\mathbb{Z}/(mn\mathbb{Z})^*\simeq \mathbb{Z}/(m\mathbb{Z})^*\times \mathbb{Z}/(n\mathbb{Z})^*$$ hence $\varphi(mn)=\varphi(m)\cdot\varphi(n)$, and if $m_1,\ldots,m_r$ are pairwise coprime, $$\varphi\left(\prod_{k=1}^{r}m_k\right)=\prod_{k=1}^{r}\varphi(m_k).$$
Alternative way: we may use the inclusion-exclusion principle to show that in general $$\varphi(n) = n\cdot\prod_{p\mid n}\left(1-\frac{1}{p}\right)$$ hence the multiplicativity of $\varphi$ follows from the fact that coprime integers have disjoint sets of prime divisors.