Let the given statement be P(n), i.e.
P (n) : (1+

1

1

)(1+

1

2

)(1+

1

3

) ... (1+

1

n

) = (n + 1)
First we prove that the statement is true for n = 1.
P (1) : (1+

1

1

) = (1 + 1) ⇒ 2 = 2 , which is true.
Assume P(k) is true for some positive integer k, i.e.,
(1+

1

1

)(1+

1

2

)(1+

1

3

) ... (1+

1

k

) = (k + 1) ... (i)
Now we shall prove that P(k + 1) is true.
For this we have to prove that
(1+

1

1

)(1+

1

2

)(1+

1

3

) ... (1+

1

k

)(1+

1

k+1

) = [(k + 1) + 1]
L.H.S. = (1+

1

1

)(1+

1

2

)(1+

1

3

) ... (1+

1

k

)(1+

1

k+1

)
= (k + 1) (1+

1

k+1

)
= (k + 1) (

k+1+1

k+1

) = 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.