Answer
Prove that $p(1)$ is true.
Prove that $p(k)$ true implies $p(k+1)$ true.
Work Step by Step
Let's note by $p(n)$ the property we want to prove:
$$p(n):\text{ } \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{n}\right)=n+1.$$
$\textbf{Step 1:}$ Prove $p(1)$ is true.
Set $n=1$ on each side and check if the property is verified:
$$\begin{align*}
1+\dfrac{1}{1}&\stackrel{?}{=}1+1\\
2&=2\checkmark.
\end{align*}$$
Therefore $p(1)$ is true.
$\textbf{Step 2:}$ Prove that if $p(k)$ is true, then $p(k+1)$ is true.
We assumed $p(k)$ is true:
$$p(k):\text{ } \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)=k+1.\tag1$$
We have to prove:
$$p(k+1):\text{ } \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=(k+1)+1.$$
$$p(k+1):\text{ } \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=k+2.\tag2$$
In Eq. $(1)$ multiply each side by $\left(1+\dfrac{1}{k+1}\right)$:
$$\begin{align*}
\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)&=(k+1)\left(1+\dfrac{1}{k+1}\right)\\
\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)&=(k+1)\left(\dfrac{k+2}{k+1}\right)\\
\text{ } \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)&=k+2.
\end{align*}$$
We got Eq. $(2)$, therefore if $p(k)$ is true, then $p(k+1)$ is also true.
Both steps being proved, the given formula is true for any natural number $n$.
