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Let $F=(X,+_X,*_X,\le_X)$ be a totally ordered field. Let $\mathbb N'$ be the isomorphic copy of natural numbers in it defined by the function $\phi:\mathbb N \rightarrow X$ such that $\phi(n)=1_X\underbrace{+_X...+_X}_{n\,times\,+}1_X$.

Let $F$ be Archimedean, i.e. $\forall x \in X\setminus{\{0_X}\}:\exists n \in \mathbb N':n|x|>1$.

How to prove that $\forall x \in X:\exists n \in \mathbb N':n>x$?

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    Case distinction, $x < 0$, $x = 0$, and $x > 0$. – 2017-01-27

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If $x>0$ then $x^{-1}\in X$ and $x^{-1}>0$ (Can you see why?). Therefore by the Archimedean property there is a $n$ s.t. $nx^{-1}>1$, thuse $n>x$.