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

Let $P(n)$ be the given statement
i.e., $P(n) : (ab)^n=a^nb^n$
We note that $P(n)$ is true for $n=1$ since $(ab)^1=a^1b^1$.
Let $P(k)$ be true, i.e.,
$\qquad (ab)^k=a^kb^k$----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Now, we have
$\qquad = (ab)^{k+1} =(ab)^k(ab)$
$\qquad = (a^kb^k) (ab)\qquad$ [ by (1) ]
$\qquad = (a^k.a^1) (b^k.b^1)=a^{k+1}.b^{k+1}$
Therefore, $P(k+1)$ is also true whenever $P(k)$ is true. Hence by principle of mathematical induction, $P(n)$ is true for all $n \in N$.