Proof
(related to Proposition: Characterization of NonArchimedean Absolute Values)
In the following prove, $(F,+,\cdot)$ is a field.
Ad $(1)$
 Assume, $\cdot$ is a nonarchimedean absolute value on $F.$
 Then $1=1$ and by the forth axiom of nonarchimedean absolute value, $2=1+1\le\max(1,1)=1,$ $3=1+1+1\le\max(2,1)=1.$
 By induction, we get $n\le 1.$
Ad $(2)$
 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.
$$(x+y)^n=\left\sum_{k=0}^n{n\choose k}x^{nk}y^k\right\le nC\max(x,y)^n,$$
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 nonarchimedean.
∎
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References
Bibliography
 Lang, Serge: "Algebra  Graduate Texts in Mathematics", Springer, 2002, 3rd Edition