We will make use of Lagrange’s Theorem: Let $G$ be a finite group and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.

Let $G=(\mathbb{Z}/n\mathbb{Z})^{\times}$ and let $H$ be the multiplicative subgroup of $G$ generated by $a$ (so $H=\{1,a,a^{2},\ldots\}$). The fact that $\gcd(a,n)=1$ ensures that $a\in G$. Notice that the order of $H$, $h=|H|$ is also the order of $a$, i.e. the smallest natural number $n>1$ such that $a^{n}$ is the identity in $G$, i.e. $a^{h}\equiv 1\mod n$. Also, recall that the order of $G=(\mathbb{Z}/n\mathbb{Z})^{\times}$ is $\phi(n)$, where $\phi$ is the Euler $\phi$ function.

By Lagrange’s theorem $h\mid|G|=\phi(n)$, so $\phi(n)=h\cdot m$ for some $m$. Thus:

as claimed. ∎