Let P(n) be the given statement
i.e., P(n):(ab)n=anbn
We note that P(n) is true for n=1 since (ab)1=a1b1.
Let P(k) be true, i.e.,
(ab)k=akbk----------(1)
We shall now prove that P(k+1) is true whenever P(k) is true.
Now, we have
=(ab)k+1=(ab)k(ab)
=(akbk)(ab) [ by (1) ]
=(ak.a1)(bk.b1)=ak+1.bk+1
Therefore, P(k+1) is also true whenever P(k) is true. Hence by principle of mathematical induction, P(n) is true for all n∈N.
