If `A` is a square matrix, using mathematical induction prove that `(A^T)^n=(A^n)^T` for all `n in N` .

Explanation

Let `P(n): (A)^(n)=(A^(n))`
`therefore P(1):(A)^(1)=(A)`
`rArr A=ArArrP(1)` is true
Now, `P(k):(A)^(n)=(A^(k))`
where `k in N`
and `P(k+1) .(A)^(k+1)=(A^(k+1))`
Where `P(k+1)` is true whenever P (k) is true.
`therefore P(k+1)(A)^(k),(A)^(1)= [A^(k+1)]`
`(A^(k)) ,(A)=[A^(k+1)]`
`(A.A^(k))=[A^(k+1)]`
`(A^(k+1))=[A^(k+1)]`