The group of complex p-power roots of unity is a proper quotient of itself

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.8
Let $p$ be a prime and let $G$ be the group of $p-$power roots of 1 in $\mathbb{C}$. Show that the map $\varphi : G \rightarrow G$ given by $z \mapsto z^p$ is a surjective group homomorphism and deduce that $G$ is isomorphic to a proper quotient of itself.

Solution: $\varphi$ is a homomorphism because $G$ is abelian. Now let $z \in G$; then $z^{p^n} = 1$ for some natural number $n$. By the fundamental theorem of algebra, the polynomial $q(x) = x^p - z$ has a root $w$ in $\mathbb{C}$. Moreover, $$(w^p)^{p^n} = w^{p^{n+1}} = 1,$$ so that $w \in G$. Finally, $\varphi(w) = z$, so that $\varphi$ is surjective.
Finally, note that again by the fundamental theorem of algebra the kernel of $\varphi$, $$\{ z \in \mathbb{C} \ |\ z^p = 1 \},$$ is nontrivial. By the First Isomorphism Theorem we have $G/\mathsf{ker}\ \varphi \cong G$.