Let the given statement be $P(n)$ i.e., $P(n)$ :
$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots \ldots .\left(1+\frac{1}{n}\right)=(n+1)$
for $n=1, \mathrm{LHS}=\left(1+\frac{1}{1}\right)=2$ and $\mathrm{RHS}=1+1=2$
$\therefore \quad P(n)$ is true for $n=1$
Let $P(k)$ is true.
i.e. $P(k)$ :
$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \ldots \ldots\left(1+\frac{1}{k}\right)=(k+1) \quad \ldots \text { (i) }$
We have to prove that $P(k+1)$ is true. Whenever $P(k)$ is true
$\therefore P(k+1)$
$\begin{aligned}
&=\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots \ldots \ldots .\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right) \\
&=(k+1)\left(1+\frac{1}{k+1}\right)=(k+1)\left(\frac{k+1+1}{k+1}\right)=k+2 \\
&=(\overline{k+1}+1)
\end{aligned}$
$\therefore \quad P(k+1)$ is also true whenever $\mathrm{P}(k)$ is true. Hence, by principle of mathematical induction $P(n)$ is true for all $n \in N$