Show that $mathbb{Z}/5mathbb{Z} = langle arangle$

Actually $mathbb{Z}/5mathbb{Z}$ is a cyclic group of order $5$, and since $5$ is prime then all the non-identity elements of $mathbb{Z}/5mathbb{Z}$ are of order $5$ then all the non-identity elements are generators of this group. Now, $a$ is an element of $mathbb{C}^*$ and a is not identity and $a^5=1$ then $|a|=5$ and $<a>$ is a cyclic group of order $5$ and having the elements ${a,a^2,a^3,a^4,1}$ where all the elements are of order $5$. So, ${a,a^2,a^3,a^4,1}$ is isomorphic to $mathbb{Z}/5mathbb{Z}$, Hence , $mathbb{Z}/5mathbb{Z}=<a>$. Is my way correct or convincing?

Denote your homomorphism by $varphi$. Using the fact that the order of $a$ is $5$, one can show that Ker $varphi$ = $5mathbb{Z}$. Now, the fundamental homomorphism theorem for groups says that $mathbb{Z}/$Ker $varphi$ $cong$ Im $varphi$. Now we have Im $varphi = ⟨a⟩$ such that $mathbb{Z}$/ $5mathbb{Z} cong ⟨a⟩$.