The following is a sentence from the proof of the theorem 1.2 (P.14-15) in Topological Vector Spaces (Third Printing Corrected 1971) by H.H.Schaefer
Finally, if $K$ is an Archimedean valuated field, then $|2| > 1$ for $2\in K$.
The following is the definition of non-Archimedean absolute value of a field in Handbook of Analysis and its foundation(P.261).
$|n e| \leq 1$ for every $n \in {\mathbb N}$,
where $e$ is the multiplicative identity of the field.
Then, I think that in an Archimedean valuated field all we can say is $|n e| > 1$ for some $n \in {\mathbb N}$. Is the statement in Topological Vector Spaces wrong ?
Assume $|2e| \le 1$, so $|2^k e| \le 1$ for all $k \ge 0$. By writing $m$ in base $2$, you get from triangle inequality
$$|me| \le 1 + \log_2(m)$$
Take $n \in \mathbb{N}$ such that $|ne| = r > 1$ and apply the inequality to $m = n^k$. For all $k \ge 0$, you have
$$r^k \le 1 + k \ \log_2(n)$$
Which becomes impossible for $k$ sufficiently large.