(ab) n = an bn for all n ϵ N

Let P (n): (ab) ⁿ = aⁿ bⁿ

Let us check for n = 1,

P (1): (ab) ¹ = a¹ b¹

: ab = ab

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): (ab) ^k = a^k b^k … (i)

We have to prove,

(ab) ^{k + 1} = a^{k + 1}.b^{k + 1}

So,

= (ab) ^{k + 1}

= (ab) ^k (ab)

= (a^k b^k) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.