Prove the rule of exponents $(ab)^{n} = a^{n}b^{n}$ by using the principle of mathematical induction for every natural number $n$.

Let $P(n)$ be the given statement:
$P(n) : (ab)^{n} = a^{n}b^{n}$
Step $1$: For $n = 1$,we have $(ab)^{1} = ab$ and $a^{1}b^{1} = ab$. Since $ab = ab$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$(ab)^{k} = a^{k}b^{k}$ .......... $(1)$
Step $3$: We need to prove that $P(k+1)$ is true,i.e.,$(ab)^{k+1} = a^{k+1}b^{k+1}$.
Consider $(ab)^{k+1} = (ab)^{k} \cdot (ab)$.
Using the assumption $(1)$,we get $(ab)^{k+1} = (a^{k}b^{k}) \cdot (ab)$.
By the associative and commutative properties of multiplication,$(ab)^{k+1} = (a^{k} \cdot a) \cdot (b^{k} \cdot b) = a^{k+1}b^{k+1}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in \mathbb{N}$.