Step 1:
Let P(n) be (A′)n=(An)′
Let P(n) be true for all n∈ N
For n=1 (A′)1=(A1)′
⇒(A)′=A′.
Hence LHS = RHS.
Hence P(n) is true for n = 1
Step 2:
Let P(n) be true for n=k.
Put n = k
(A′)k=(Ak)′
Multiply A′ on both the side
(A′)k.A′=(Ak)′.A′
⇒ A′k.A′ = (Ak)′.A′
⇒ A′k.A′1 = (Ak.A1)′
⇒ (A′)k+1 = (Ak+1)′
Hence P(n) is true for n = k+1.