Induction proof of exponent properties

Question

I have the following induction proof and need a verification.I do not know if this is the correct way to approach this problem.Any ideas?

I need to prove that if $0<x<1$,then for all $n \in \mathbb{N}$ with $n>1$, $x^n<x$.

Proof attempt:

Base Case Since $0<x<1$, $x^{-1}>1$, then $(x^2)(x^{-1})>x^2 \implies x^2<x$

Inductive step Assume for $k$, $x^k<x$ with $0<x<1$ prove $x^{k+1}<x$

we have $(x)(x^k)<(x)(x)<x \implies x^{k+1}<x$ hence true for all naturals

Answer 1

In the base case: Why switch the sense of the inequalities. Doing so at the implication is less clear than not doing so. Also, working backwards from the conclusion to order each inequality in the chain gives

$1 < x^{-1}$, then $x^2 < x^{-1}x^2 = x$, as desired.

Inductive step: "Assume for $k$" ... what? $>0$? $<0$? odd? even? I suspect you mean "Assume for any natural number $k > 2$, ...". I explicitly mention the natural numbers because that is the set of exponents about which you are proving something.

Also, I probably would insert $$ x^{k+1} = x \cdot x^k < x \cdot x = x^2 < x $$ to more explicitly tie both the inductive target and the base case to the chain of relations in the inductive step.