Prove the following by using the principle of mathematical induction for all n ∈ N:

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

$\mathrm{P}(n):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)=(n+1)$

For n = 1, we have

$P(1):\left(1+\frac{1}{1}\right)=2=(1+1)$, which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$\mathrm{P}(k):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{k}\right)=(k+1)$ $\ldots(1)$

We shall now prove that P(k + 1) is true.

Consider

$\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right)$

$=(k+1)\left(1+\frac{1}{k+1}\right) \quad[$ Using $(1)]$

$=(k+1)\left(\frac{(k+1)+1}{(k+1)}\right)$

$=(k+1)+1$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.