Prove the following by using the principle of mathematical induction for all $n \in N$:
$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{n}\right)=(n+1)$

Let the given statement be $P(n)$,i.e.,
$P(n): \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{n}\right)=(n+1)$
For $n=1$,we have
$P(1): \left(1+\frac{1}{1}\right) = 2 = (1+1)$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$P(k): \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{k}\right)=(k+1)$ .........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right)$
$= (k+1)\left(1+\frac{1}{k+1}\right)$ [Using $(i)$]
$= (k+1)\left[\frac{(k+1)+1}{k+1}\right]$
$= (k+1)+1$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.