Proof

In the following prove, $(F,+,\cdot)$ is a field.

Assume, $|\cdot|$ is a non-archimedean absolute value on $F.$
Then $|1|=1$ and by the forth axiom of non-archimedean absolute value, $|2|=|1+1|\le\max(|1|,|1|)=1,$ $|3|=|1+1+1|\le\max(|2|,|1|)=1.$
By induction, we get $|n|\le 1.$

Assume, there is a constant $C$ with $|n|\le C,$ where $n=\underbrace{1+\ldots+1}_{n\text{ times}},$ $1\in F.$
Then, for fixed $x,y\in F,$ we get with the binomial theorem. $$|(x+y)^n|=\left|\sum_{k=0}^n{n\choose k}x^{n-k}y^k\right|\le nC\max(|x|,|y|)^n,$$ in which, by abuse of notation, we regard $n$ also as a natural number.
Taking $n$-th roots on both sides a letting $n$ tend to infinity, we get by the standard results from analysis (limit of nth root of n and limit of the nth root of a positive constant), we get $$|x+y|\le\max (|x|,|y|).$$
It follows that $|\cdot|$ is non-archimedean.
∎

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Lang, Serge: "Algebra - Graduate Texts in Mathematics", Springer, 2002, 3rd Edition