Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=2$ and $RHS=2$, thus it is true for $n=1$.
2. Assume it is true for $n=k$, we have $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{k})=k+1$
3. For $n=k+1$, we have $LHS=(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{k})(1+\frac{1}{k+1})=(k+1)(1+\frac{1}{k+1})=k+1+1=k+2=RHS$
4. The statement is true for any natural number $n$.
