Lemma. \[\sum_{d|n} \phi(d) = n.\]
Proof. By induction: \[\sum_{d|p^{k}m}\phi(d) = \sum_{i=0}^{k}\sum_{d|m}\phi(p^{i}d) = \sum_{i=0}^{k}\sum_{d|m}\phi(d) \phi(p^{i}).\] But this is telescoping because \(\phi(p^{i})=p^{i}-p^{i-1}\). It tellescopes to \[p^{k}\sum_{d|m}\phi(d) = p^{k}m,\] by induction.
Remark. Actually, we can prove this a nicer way. Partition \([n]\) based on \(\gcd(x,n)\). For each \(d\mid n,\) the number of \(x\) with \(\gcd(x,n) = d\) is \(\phi(n/d)\). Thus we have \[n = \sum_{d\mid n}\phi(n/d)=\sum_{d\mid n}\phi(d).\]
Theorem. The multiplicative group of \(\mathbb{F}_p\) is cyclic.
In fact, this is true of any finite field by a similar proof.
Proof. For each \(d|p-1\), there are at most \(\phi(d)\) elements of \(\mathbb{F}_p\) which have order \(d\). Thus, by the lemma above, for each \(d|p-1\) there are exactly \(\phi(d)\) elements of \(\mathbb{F}_p\) with order \(d\). In particular, \(\phi(p-1)>0\), so there is some element with order \(p-1\), i.e. a generator.