If $AB = BA$ for any two square matrices, then prove by mathematical induction that ${(AB)^n} = {A^n}{B^n}$.

Let $P(n):{(AB)^n} = {A^n}{B^n}$
$\therefore$ $P(1):{(AB)^1} = {A^1}{B^1} \Rightarrow AB = AB$

Therefore, $P(1)$ is true.
Now, $P(k):{(AB)^k} = {A^k}{B^k},k \in N$

Therefore, $P(K)$ iș true, whenever $P(k + 1)$ is true.

$\therefore$ $P{(K + 1:AB)^{k + 1}} = {A^{k + 1}}{B^{k + 1}}$

…..(i)
$\Rightarrow$ $A{B^K} \cdot A{B^1}$

$\Rightarrow$ ${A^k}{B^k} \cdot BA \Rightarrow {A^k}{B^{k + 1}}A$

$\Rightarrow$ ${A^k} \cdot A \cdot {B^{k + 1}} \Rightarrow {A^{k + 1}}{B^{k + 1}}$

$\Rightarrow$ ${(A \cdot B)^{k + 1}} = {A^{k + 1}}{B^{k + 1}}$

Therefore, $P(k + 1)$ is true for all $n \in N$, whenever $P(k)$ is true.

By mathematical induction $(AB) = {A^n}{B^n}$ is true for all $n \in N$.