I am working on the following question.
Let G be a group and let $n$ be any integer. For each integer $n$ define the map $\phi_n:G \rightarrow G$ by $\phi_n(a)=a^n$ for every element $a$ in the group. Prove that if $G$ is abelian then $\phi_n$ is a homomorphism for any $n \in \mathbb{Z}$.
Note: I have seen similar proofs on this forum for related questions when $n=2$.
My work thus far is:
Let $a,b \in G$. Consider $\phi_n(ab)=(ab)^n=\underbrace{ab \cdot ab \cdot ab \cdots ab}_{n}=\underbrace{a \cdot a \cdots a}_{n} \cdot \underbrace{b \cdot b \cdots b}_{n}=a^n \cdot b^n=\phi_n(a) \phi_n(b)$. Where $(ab)^n=a^n b^n$ is true since $G$ is abelian.
This seems straightforward except I feel as though more steps are needed to explain why $\underbrace{ab \cdot ab \cdot ab \cdots ab}_{n}=\underbrace{a \cdot a \cdots a}_{n} \cdot \underbrace{b \cdot b \cdots b}_{n}$ is true other than by saying because $G$ is abelian.
Thank you.