Precalculus (6th Edition) Blitzer

Step 1. Checking for $n=1$, we have $0\lt x \lt 1$, which is true. Step 2. Assume the statement is true for $n=k$; we have $0\lt x^k \lt 1$ Step 3. For $n=k+1$, as $0\lt x\lt 1$, multiply $x$ to the above inequalities. We have $0\cdot x\lt x\cdot x^k\lt 1\cdot x$, which leads to $0\lt x^{k+1} \lt x \lt 1$. Thus, the statement is also true for $n=k+1$ Step 4. Thus, through mathematical induction, we proved that the statement is true for every positive integer n.