(ab)ⁿ = aⁿbⁿ for all n ∈ N.

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Solution

Let P(n) be the given statement.
Now,
$P (n) : (a b)^{n} = a^{n} b^{n} for all n \in N . Step 1 : P (1) : (a b)^{1} = a^{1} b^{1} = a b Thus, P (1) is true . Step 2 : Let P (m) be true . Then, (a b)^{m} = a^{m} b^{m} We need to show that P (m + 1) is true whenever P (m) is true . Now, P (m + 1) : (a b)^{m + 1} = (a b)^{m} . a b = a^{m} b^{m} . a b = a^{m} a . b^{m} b = a^{m + 1} b^{m + 1} Hence, P (m + 1) is true . B y t h e p rinciple of mathematical i nduction, P (n) is true for all n \in N .$

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(ab)n = anbn for all n ∈ N.

(ab)ⁿ = aⁿbⁿ for all n ∈ N.