Is it possible to show that $\phi$ is multiplicative using Euler's theorem? That is, can I show
Theorem. Let $m$ and $n$ be relatively prime positive integers. Then $\phi(mn)=\phi(m)\phi(n)$
using
Euler's Theorem. If $m$ is a positive integer and $a$ is an integer with $(a,m)=1$, then $a^{\phi(m)}\equiv1\pmod{m}$?
I have tried a few things, but since they are all congruences, I do not know whether they would be general enough to show this or not.
The idea that I have is the following:
Let $(x,ab)=1$ and $(a,b)=1$. Then $\begin{align*} x^{\phi(ab)}&\equiv 1\pmod{ab},\\ a^{\phi(b)}&\equiv 1\pmod{b},\\ \text{and }\quad b^{\phi(a)}&\equiv 1\pmod{a}. \end{align*}$
Taking the logarithms of these expressions, we obtain: $\begin{align*} \phi(ab) &\equiv 0 \pmod{ab},\\ \phi(b) &\equiv 0 \pmod{b},\\ \text{and }\quad \phi(a)&\equiv 0\pmod{a}. \end{align*}$
Multiplying the last two expressions by $a$ and $b$, respectively, we obtain:
$a\phi(b)\equiv0\pmod{ab},\quad \text{and}\quad b\phi(a)\equiv0\pmod{ab},$ and multiplying these two: $ab\phi(a)\phi(b)\equiv\phi(a)\phi(b)\equiv0\pmod{ab}.$ Can I then conclude that since $\phi(ab)\equiv0\pmod{ab}\text{ and } \phi(a)\phi(b)\equiv0\pmod{ab},$ then $\phi(ab)=\phi(a)\phi(b)?$