Prove of some property of square matrix

For square matices A and B

AB = BA and ABⁿ=BⁿA

Let P(n): (AB)ⁿ=A ⁿBⁿ

For n=1,

L.H.S = AB and R.H.S =AB

 P(n) is true for 1.

Let P(n) be true for n = k.

(AB)^k = A^kB^k

Multiply both the side by AB

L.H.S = (AB)^k(AB) =(AB)^k+1

R.H.S = A^kB^k (AB) = A^kB^k (BA)             [ AB = BA]

           = A^k (B^k B) A

           = A^k B^k+1  A              [ABⁿ=BⁿA ]

           = A^k  (A B^k+1)  

           = A^k+1 B^k+1

⇒ P(n) is true for n = k+1

By principle of mathematical induction

P(n) is true for n N.

Hence, (AB)ⁿ = AⁿBⁿ