Use induction to show that $(a\cdot b)^n = a^n \cdot b^n$ for any $n \geq 1$.
Show that $a^{-n} = (a^{-1})^n$ and then prove it for negative integers.
Solution: Since $G$ is an abelian group, for any $a,b \in G$, we have $a\cdot b = b \cdot a$. We need to prove hat for any $n \in \mathbb{Z} $, $(a\cdot b)^n = = a^n \cdot b^n$. Let us first prove this for $n \in \mathbb{N} $. We will prove this by the method of induction. Let $P(n)$ be the statement that $(a\cdot b)^n = a^n \cdot b^n$.
Since $G$ is abelian, $P(1)$ is true.
Let $P(k)$ is true.
We need to prove that $P(k+1)$ is true. Note that
\begin{align*}
(a\cdot b)^{k+1} & = (a\cdot b)^k \cdot (a\cdot b) = (a^k \cdot b^k ) \cdot a \cdot b \\
& = a^k \cdot (b^k \cdot a) \cdot b = a^k \cdot (a \cdot b^k) \cdot b \\
& = a^{k+1} \cdot b^{k+1}.
\end{align*}
Thus by the method of induction, we conclude that for $n \in \mathbb{N} $, $(a \cdot b)^n = a^n \cdot b^n$.
For $n\in \mathbb{N} $, we note that
\begin{gather*}
(a\cdot b)^{-1} = b^{-1} \cdot a^{-1} = a^{-1} \cdot b^{-1} \text{ and } \\
(a^{-1} )^n = a^{-n}.
\end{gather*}
Thus, using the above two, we have
\begin{align*}
(a\cdot b)^{-n} = \left( (a\cdot b)^{-1} \right) ^n = \left( a^{-1} \cdot b^{-1} \right) ^n = a^{-n} \cdot b^{-n}.
\end{align*}
Therefore, for $n \in \mathbb{Z} $, we proved that $(a\cdot b)^n = a^n \cdot b^n$.