A finite field cannot be an ordered field.

I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to $\mathbb R$. Since all ordered finite fields would have supremum property that must mean none exist. Could someone please show me a proof of this?

Thank you very much, Regards.

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HINT: Suppose that $(F,0,1,+,\cdot,<)$ is an ordered field which is finite of characteristic $p$. Then $0<1<1+1<\ldots$, conclude a contradiction.

Hint $\$ In an ordered ring, positives are closed under addition (so a sum of positives is $\ne 0$).

Remark $\$ More generally, note that linearly ordered groups are torsion free: $\rm\: 0\ne n\in \mathbb N,$ $\rm\:g>0 \:\Rightarrow\: n\cdot g = g +\cdots + g > 0,\:$ since positives are closed under addition. Conversely, a torsion-free commutative group can be linearly ordered (Levi, $1942$).

Hint: any finite field must have a non-zero characteristic.