Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: 0\lt x^1=x\lt1$ , thus this is true by the condition.
2) Assume for $n=k: 0\lt x^k\lt1$. Then for $n=k+1:0\lt x^{k+1}=x\cdot x^k\lt1.$ This statement is true because we multiplied the value in the middle of the inequality of the inductive hypothesis by a number larger than $0$ (thus it stayed positive) but by a number less than $1$, thus it became smaller thus it is still less than $1$. Thus we proved what we wanted to.