What's the proof that the Euler totient function is multiplicative?

In general, if $R$ and $S$ are rings, then $R\times S$ is a ring. The units of $R\times S$ are the elements $(r,s)$ with $r$ a unit of $R$ and $s$ a unit of $S$. If $R$ and $S$ are finite rings, the number of units in $R\times S$ is therefore the number of units in $R$ times the number of units in $S$.

Now if $\gcd(A,B)=1$, then $$\mathbb Z/\left<AB\right> \cong \mathbb Z/\left<A\right> \times \mathbb Z/\left<B\right>$$

This is essentially the Chinese remainder theorem.

But the number of units in the ring $\mathbb Z/\left<n\right>$ is $\phi(n)$. So the number of units in $\mathbb Z/\left<AB\right>$ is $\phi(AB)$ and the number of units in $\mathbb Z/\left<A\right> \times \mathbb Z/\left<B\right>$ is $\phi(A)\phi(B)$

$$n = \prod_{k=1}^{z}p_{k}^{e_{k}}$$ $$\varphi(n) = n \prod_{k=1}^{z}(1-\frac{1}{p_{k}})$$ Let $$a=\prod_{k=1}^{w} p_{k}^{e_{k}}\quad b=\prod_{k=w+1}^{z} p_{k}^{e_{k}}$$ $n=ab$, then $$\varphi(a) = a\prod_{k=1}^{w}(1-\frac{1}{p_{k}})$$ $$\varphi(b) = b\prod_{k=w+1}^{z}(1-\frac{1}{p_{k}})$$ $$\varphi(a)\varphi(b) = ab\prod_{k=1}^{w}(1-\frac{1}{p_{k}}) \cdot \prod_{k=w+1}^{z}(1-\frac{1}{p_{k}})$$ $$=ab\prod_{k=1}^{z}(1-\frac{1}{p_{k}})$$ $$=n\prod_{k=1}^{z}(1-\frac{1}{p_{k}})$$ $$=\varphi(n)$$ $$\varphi(a)\varphi(b)=\varphi(ab)$$

If $\phi(n)$ is the Euler's Totient Function, then the proof goes as follows :

By definition $\phi(p)=p-1$ if p is prime,

Now, if $n$ is any composite number, then $n=pq$, where $p$ and $q$ are prime.

To see that $\phi(n)=\phi(p)\times \phi(q)$, consider that the set of positive integers less than $n$ is the set $\{1,.....(pq-1)\}$. The integers in this set that are not relatively prime to $n$ are the set $\{p,2p,....(q-1)p\}$and the set $\{q,2q,....(p-1)q\}$. Accordingly

Then $\phi(n)=\phi(pq)=pq-1-[(q-1)+(p-1)]=pq-(p+q)+1=(p-1)(q-1)=\phi(p)\phi(q)$ as required . $\square$