Answer
See explanations.
Work Step by Step
Step 1. Prove that the formula is true for $n=1$: $LHS=1+\frac{1}{1}=2$, $RHS=1+1=2$, thus it is true for $n=1$
Step 2. Assume the formula is true for $n=k$: we have $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{k})=k+1$
Step 3. Prove that it is also true for $n=k+1$:
$LHS=(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{k})(1+\frac{1}{k+1})=(k+1)(1+\frac{1}{k+1})=k+1+1=k+2$
$RHS=k+1+1=k+2=LHS$
Thus, the formula is also true for $n=k+1$
Step 4. With mathematical induction, we have proved that the formula is true for all natural numbers n.
