Let the given statement be P(n), i.e. P (n) : $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ ... $\left(1+\frac{1}{n}\right)$ = (n + 1) First we prove that the statement is true for n = 1. P (1) : $\left(1+\frac{1}{1}\right)$ = (1 + 1) ⇒ 2 = 2 , which is true. Assume P(k) is true for some positive integer k, i.e., $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ ... $\left(1+\frac{1}{k}\right)$ = (k + 1) ... (i) Now we shall prove that P(k + 1) is true. For this we have to prove that $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ ... $\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)$ = [(k + 1) + 1] L.H.S. = $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ ... $\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)$ = (k + 1) $\left(1+\frac{1}{k+1}\right)$ = (k + 1) $\left(\frac{k+1+1}{k+1}\right)$ = k + 2 = [(k + 1) + 1] = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.