Euler's theorem can be proved as follows: Take $r_{1}, r_{2}, \cdots, r_{\phi(m)}$ as a reduced residue system modulo $m$. Since $(a, m)=1$, it follows that $a r_{1}, a r_{2}, \cdots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$ (see Unit 6). Because two complete (reduced) residue systems modulo $m$ are permutations of each other modulo $m$, we have in particular
$$r_{1} \cdots r_{\phi(m)} \equiv a r_{1} \cdot a r_{2} \cdots \cdot a r_{\phi(m)}=a^{\phi(m)} r_{1} r_{2} \cdots r_{\phi(m)}(\bmod m)$$

Since $\left(r_{i}, m\right)=1$, it follows that $\left(r_{1} r_{2} \cdots r_{\phi(m)}, m\right)=1$, so the term $r_{1} \cdots r_{\phi(m)}$ can be canceled from both sides of the above equation, yielding $a^{\phi(m)} \equiv 1(\bmod m)$.