Show that, if $n\in\mathbb{Z}$ and $G$ is an Abelian group, $(ab)^n=a^nb^n$ for all $a$ and $b$ in $G$.
Proof:
For positive $n$, we know the statement is true for $n=1$ by $(ab)^1=ab=a^1b^1=ab\space\checkmark$. Now we will proceed by induction on $n$, assuming the statement is true for $n=k$ to show it is true for $n=k+1$.
Letting $z=ab$,
$$\begin{align} z^{k+1}&=z^kz^1\\ &=(ab)^kz\\ &=(a^kb^k)z\\ &=a^kb^kab\\ &=a^kb^kba \\ &=a^kb^{k+1}a \\ &=aa^kb^{k+1} \\ &=a^{k+1}b^{k+1}.\checkmark \end{align}$$
For negative $n$, we know the statement is true for $n=-1$ since $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$ since $G$ is Abelian. Now we will proceed by induction on $n$, assuming the statement to be true for $n=-k$ to show that the statement is true for $n=-k-1$.
Letting $z=ab,$
$$\begin{align} z^{-k-1}&=z^{-k}z^{-1}\\ &=(ab)^{-k}z^{-1}\\ &=(a^{-k}b^{-k})(ab)^{-1} \\ &=(a^{-k}b^{-k})(b^{-1}a^{-1}) \\ &=(a^{-k}b^{-k-1})a^{-1} \\ &=a^{-1}(a^{-k}b^{-k-1}) \\ &=a^{-k-1}b^{-k-1}.\space\checkmark \end{align}$$
Hence the statement is true for all integers.
Is this a Valid proof? Thanks!
As @Namaste points out in the comments . . .
Yes, you proof is valid, provided you mention that the group $G$ is abelian in all the right places.