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

THM Assume that $(a,b)=1$. Then $$(a,y)=1 \text{ and } (b,x)=1\iff (ax+by,ab)=1$$

P We prove the contrapositive of each direction.

$(\Rightarrow)$ Suppose thus that there is a prime $p$ such that $p\mid (ax+by,ab)$. Then $p\mid ab$. Without loss of generality, assume $p \mid a$. Since $p\mid ax+by$, we have $p\mid by$, and since $(a,b)=1$, $p\mid y$. Thus $p\mid (a,y)\implies (a,y)>1$. We have thus proven, under the hypothesis that $(a,b)=1$; that $$(ax+by,ab)>1\implies (a,y)>1 \text{ or } (b,x)>1$$ since the other option would have been assuming that $p\mid a$.

$(\Leftarrow)$ Now suppose $(x,b)>1$. Then $(ax+by,ab)>1$ since $(x,b)\mid ax+by$. Analogously, $(a,y)>1$ implies $(ax+by,ab)>1$. $\blacktriangle$

COR Suppose that $(a,b)=1$, and that $x$ ranges through the $\phi(b)$ numbers coprime to $b$ and $y$ ranges throughout the $\phi(a)$ numbers coprime to $a$. Then $ax+by$ ranges throughout the $\phi(a)\cdot\phi(b)$ numbers coprime to $ab$, which in turn is $\phi(a\cdot b)$.