Show that the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic

I'm looking for a proof of this that does not rely on the Chinese remainder theorem, since the exercise is from a book that has not reached such a point (Aluffi, Algebra).

Show that for $p$ prime, the multiplicative group $G=(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic. You may use that for $p$ prime, the equation $x^d=1$ has at most $d$ solutions in $\mathbb{Z}/p\mathbb{Z}$. (hint: for $g\in G$ of maximal order, $\forall h\in G, |h|\;|\;|g|$, and in particular, $h^{|g|}=1$.)

My only idea for now is constructing an element $a\in G$ such that $|a|=|G|$, using the fact that a group of order $n$ is cyclic $\iff\exists$ an element of order $n$.

If you know the structure theorem for finite Abelian groups, you will be able to prove that either a finite Abelian group $G$ is cyclic, or that its exponent is $<|G|$, that is there is $m<|G|$ with $g^m=e$ for for all $g\in G$.

An alternative attack: show that for $d\mid (p-1)$ the group $G=(\Bbb Z/\Bbb Z)^*$ has exactly $d$ solutions of $x^d\equiv1$. If you compare $G$ to $H=\Bbb Z/(p-1)\Bbb Z$, you find that the number of elements of order dividing $d$ is the same in each, therefore the number of elements of order exactly $d$ is the same in each, and now take $d=p-1$.

Hint: Use the hint.

Among all elements if $(\Bbb Z/p\Bbb Z)^\times$, let $g$ have maximal order, $d$. If we assume that $d<p-1$, then there exists $h\in (\Bbb Z/p\Bbb Z)^\times$ that is not among the $d$ solutions of $x^d=1$. The order $m$ of $h$ is thus not a divisor of $d$. Let $h'=h^{m/\gcd(m,d)}$. Then $h'$ is of order $m'=\frac m{\gcd(m,d)}$, which is coprime to $d$. Conclude that $hg$ has order $m'd>d$, contradiction.

Let $d\leq p-1$ be the maximal order. Now since the order of an element in $(\mathbb{Z}/p\mathbb{Z})^*$ divides $d$ there are $p-1$ solutions of $x^d=[1]$. So we have $p-1\leq d$, and $d=p-1$.