If $ \langle G, \star \rangle $ is an abelian group, then for all $a, b \in G$, show that $(a \star b)^{n} = a^{n} \star b^{n}$. If $ \langle G, \star \rangle $ is an abelian group, then for all $a, b \in G$, show that $(a \star b)^{n} = a^{n} \star b^{n}$.
I am stuck at the first step, unable to figure out how to start. I am just beginning on group theory, and know that an abelian group obeys commutativity, associativity, has an identity element, and each element is invertible. I am unable to figure out how to apply this for the proof. Any hints on how I should get started?
 A: Use induction on $n$; I'll give an outline of the induction step, and let you fill in all the relevant assumptions, details, and computations.
\begin{align*}
(a \star b)^{n + 1} &= (a \star b)^n \star (a \star b) \\
&= a^n \star b^n \star a \star b \\
&= a^n \star a \star b^n \star b \\
&= a^{n + 1} \star b^{n + 1}
\end{align*}
Be sure that you can justify all the steps used above.
A: $$(a \star b)^n = a \star b \star a \star  b \star a \star b \cdots
= a\star a \star b \star b \star a \star b \cdots ~~~ \mbox{interchanging one $a$ and $b$}$$
Now do this repeatedly to get all the $a$'s to the left.
Or use induction to prove it
A: Hint: $$(a\star b)^n = \underbrace{(a\star b)\star \dots \star(a\star b)}_{\text{$n$ copies}}$$
$$a^n\ast b^n = \underbrace{a\star\dots\star a}_{\text{$n$ copies}}\star\underbrace{b\star\dots\star b}_{\text{$n$ copies}}$$
Note that you can drop the brackets in the first line due to associativity of $\star$. You should be able to use the fact that $\star$ is commutative to get from the first line to the second.
