## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Since the given statement is true for $n=1$, and that when assumed to be true for a natural number k, the given statement is also true for $n=k+1$, then by the principle of Mathematical Induction, the given statement is true for all natural numbers $n$. (Refer to the step-by-step work below for the actual proof) Thus, for all natural numbers $n$, if $x \gt 1$, then $x^n \gt 1$.
a) The statement must hold for the first natural number, known as the base case. b) Assume that the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we will prove that the statement also holds for $n + 1$. This is known as the inductive step. So, we have: For $n=1$: when $x\gt1$, then, we have: $x^1=x\gt1$ Assume for $n=k$ ; $x \gt 1$, then $x^k\gt1$ Then for $n=k+1$: $x^{k+1} =(x^k)( x)$ Since $x^k>1$ and $x>1$, then $x^k(x)>1$. Thus we can see that the above two conditions are satisfied. So, by the Principal of Mathematical Induction, the given statement has been proved true for all natural numbers.