Prove that (ab)n=anbn(ab)n=anbn is true for every natural number n

Prove $(ab)^n=a^nb^n$ is true for every natural number $n$ by induction.

Let $P(n)$ be the proposition that

$(ab)^n=a^nb^n$

for the natural number $n.$

BASIS STEP:

$P(1)$ ​is true because $(ab)^1=ab=a^1b^1.$

This completes the basis step.

INDUCTIVE STEP: For the inductive hypothesis, we assume that $P(k)$ is true for an arbitrary natural number $k.$ That is, we assume that

$(ab)^k=a^kb^k$

To carry out the inductive step using this assumption, we must show that when we assume that $P(k)$ is true, then $P(k+1)$ is also true. That is, we must show that

$(ab)^{k+1}=a^{k+1}b^{k+1}$

assuming the inductive hypothesis $P(k).$ Under the assumption of $P(k),$ we see that

$(ab)^k=a^kb^k$

$(ab)(ab)^k=(ab)a^kb^k$

$(ab)^{k+1}=aba^{k}b^{k}$

$(ab)^{k+1}=aa^{k}bb^{k}$

$(ab)^{k+1}=(aa^{k})(bb^{k})$

$(ab)^{k+1}=a^{k+1}b^{k+1}$

We have completed the inductive step.

Because we have completed the basis step and the inductive step, by mathematical induction we know that $P(n)$ is true for all natural numbers $n.$

That is, by principle of mathematical induction

$(ab)^n=a^nb^n$

for all natural numbers $n$