Prove by mathematical induction that ${\left( {{A^\prime }} \right)^n} = {\left( {{A^n}} \right)^\prime }$ where $n \in N$ for any square matrix A.

Let $P(n):{\left( {{A^\prime }} \right)^n} = {\left( {{A^n}} \right)^\prime }$

$\therefore$ $P(1):{(A)^1} = {(A)^\prime }$

$\Rightarrow$ ${A^\prime } = {A^\prime } \Rightarrow P(1)$

is true
Now, $P(k):{\left( {{A^\prime }} \right)^k} = {\left( {{A^k}} \right)^\prime }$

where $k \in N$
and $P(k + 1):{\left( {{A^\prime }} \right)^{k + 1}} = {\left( {{A^{k + 1}}} \right)^\prime }$

where $P(k + 1)$ is true

whenever $P(k)$ is true.

$\therefore$ $P(k + 1):{\left( {{A^\prime }} \right)^k} \cdot {\left( {{A^\prime }} \right)^1} = {\left[ {{A^{k + 1}}} \right]^\prime }$

${\left( {{A^k}} \right)^\prime } \cdot {(A)^\prime } = {\left[ {{A^{k + 1}}} \right]^\prime }$

${\left( {A \cdot {A^k}} \right)^\prime } = {\left[ {{A^{k + 1}}} \right]^\prime }$

${\left( {{A^{k + 1}}} \right)^\prime } = {\left[ {{A^{k + 1}}} \right]^\prime }$