Following the hint, we prove the equalities by induction on $m$ , but we prove them for all $n\in\mathbb{Z}_{+}$ at the same time, why not? Let $A$ , $B$ and $C$ be the sets of all positive integers $m$ such that the first, second and third equality above, respectively, holds for every $n\in\mathbb{Z}_{+}$ and $a,b\in\mathbb{R}$ . Then, $1\in A\cap B\cap C$ , as, by definition, $x^{1}=x$ for all $x\in\mathbb{R}$ , and for all $n\in\mathbb{Z}_{+}$ and $a,b\in\mathbb{R}$ , $a^{n}\cdot a^{1}=a^{n}\cdot a=a^{n+1}$ , $(a^{n})^{1}=a^{n}=a^{n\cdot1}$ , and $a^{1}\cdot b^{1}=ab=(ab)^{1}$ . Now, we show that if $m\in A$ , then $m+1\in A$ . Again, by definition, for every $n\in\mathbb{Z}_{+}$ and $a\in\mathbb{R}$ , $a^{n}a^{m+1}=a^{n}\cdot(a^{m}\cdot a)$ $=(a^{n}\cdot a^{m})\cdot a$ $=a^{n+m}a$ $=a^{(n+m)+1}$ $=a^{n+(m+1)}$ . Hence, $A$ is inductive, and $A=\mathbb{Z}_{+}$ . We are going to use this fact to prove that $B$ is inductive as well. By definition, for every $n\in\mathbb{Z}_{+}$ and $a\in\mathbb{R}$ , $(a^{n})^{m+1}=(a^{n})^{m}a^{n}$ $=a^{nm}a^{n}$ [here we use the fact that $A=\mathbb{Z}_{+}$ ] $=a^{nm+n}$ $=a^{n(m+1)}$ , hence, $B=\mathbb{Z}_{+}$ . Finally, again by definition, for every $n\in\mathbb{Z}_{+}$ and $a,b\in\mathbb{R}$ , $a^{m+1}b^{m+1}=(a^{m}a)(b^{m}b)$ $=(a^{m}b^{m})(ab)$ $=(ab)^{m}(ab)$ $=(ab)^{m+1}$ , hence, $C$ is inductive, and $C=\mathbb{Z}_{+}$ .