Prove $a^{n+1} < a$ if and only if $a < 1$ for all $a > 0$ in any ordered field

Question

Prove $a^{n+1} < a$ if and only if $a < 1$ for all $a > 0$ in any ordered field.

Proof: We first prove that for $a, b > 0$, then $ab < b$ if and only if $a < 1$.

If $a > 1$, then $a - 1 > 0$, so $(a - 1)b > 0$, and $ab > b$. Conversely, if $a < 1$, $ab < b$. And clearly if $a = 1$, then $ab = b$.

By induction, we conclude $a^{n+1} < a$ if and only if $a < 1$.

Questions:

1. Is this proof correct? Rigorous? Well-written? How can it be improved?
2. I'm surprised that I need to use addition and the distributive property for what is essentially a result concerning multiplication only. Is there a proof that avoids addition and distribution? Or is the claim inherent to the field aspects? Put in other words: Would an ordered group have a similar property?

Answer 1

I think we need $n\geq 1$ for this proposition. I'll try to answer your second question.

$(\Rightarrow)$ Suppose that $a\geq 1$, we have two possibilities: If $a=1$ then $a^{n+1}=a$; if $a>1$ then $a²>a>1$ wich implies $a²>1$ and by induction we have $a^n>1$ and then $a^{n+1}>a$. Both possibilities contradicts the hypotesis.

$(\Leftarrow)$ Now, we have $0<a<1$ by hipothesis. Multiply $a$ in the inequality we have $0<a²<a<1\Rightarrow 0<a²<1$; By induction, we have $0<a^n<1$ then multiply $a$ again and we have $0<a^{n+1}<a$.

(Sorry for my bad english)