Appendix - Mathematical Induction and Other Forms of Proofs - Exercises - Page A6: 4

See the proof below.

In order to show that the formula holds we must show that it is true for $n=1$ and that if its is true for $n=k$ then, it is true for $n=k+1$. $\textbf{1.}$ The formula holds for $n=1$ since $$\left(1+\displaystyle\frac{1}{1}\right)=1+1.$$ $\textbf{2.}$ Suppose the formula holds for $n=k$, that is, suppose that $$\left(1+\displaystyle\frac{1}{1}\right)\left(1+\displaystyle\frac{1}{2}\right)\left(1+\displaystyle\frac{1}{3}\right)\ldots\left(1+\displaystyle\frac{1}{k}\right)=k+1.$$ Then, multiplying both sides of this equality by $\left(1+\displaystyle\frac{1}{k+1}\right)$ gives us that $$\left(1+\displaystyle\frac{1}{1}\right)\left(1+\displaystyle\frac{1}{2}\right)\left(1+\displaystyle\frac{1}{3}\right)\ldots\left(1+\displaystyle\frac{1}{k}\right)\left(1+\displaystyle\frac{1}{k+1}\right)=\left(k+1\right)\left(1+\displaystyle\frac{1}{k+1}\right).$$ Now, observe that $$\left(k+1\right)\left(1+\displaystyle\frac{1}{k+1}\right)=\left(k+1\right)+\left(k+1\right)\left(\displaystyle\frac{1}{k+1}\right)=\left(k+1\right)+1.$$ Thus, $$\left(1+\displaystyle\frac{1}{1}\right)\left(1+\displaystyle\frac{1}{2}\right)\left(1+\displaystyle\frac{1}{3}\right)\ldots\left(1+\displaystyle\frac{1}{k}\right)\left(1+\displaystyle\frac{1}{k+1}\right)=\left(k+1\right)+1$$ and therefore, the formula holds for $n=k+1$ whenever it is true for $n=k$. Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$.