If $k$ is an ordered field, the least ordinal $s(k)$ which doesn't embed in $(k,<)$ is regular. This is because every interval of an ordered field embeds in every infinite interval so given a strictly increasing map $f: \alpha < s(k) \rightarrow s(k)$ you can define an embedding $(\sup(f(\alpha)),\in) \rightarrow (k,<)$. Moreover $s(k)$ is always uncountable since $s(\mathbb{Q}) = \omega_1$.

Note that these results make use of the field structure of $k$. They fail even for ordered rings such as the ring obtained from $({\omega}^{\omega},\underline{+},\underline{\times})$ (natural operations), by adding additive inverses and extending the operations like one usually does.

Let $\lambda$ be an $\varepsilon$-number. Let $No(\lambda)$ denote the field of surreal numbers of length $<\lambda$. I am trying to figure out what $s(No(\lambda))$ is.

Clearly, $s(No(\lambda))$ is at least equal to ${\lambda}^+$ and at most equal to ${|No(\lambda)|}^+$. Assuming GCH, this yields $s(No(\lambda)) = {\lambda}^+$ when $\lambda$ is itself a cardinal, and ${\lambda}^+ \leq s(No(\lambda)) \leq ({\lambda}^+)^+$ otherwise. This is just a computation of $|No(\lambda)|$ using elemenraty cardinal arithmetic and the ordered-set-focused definition of $No(\lambda)$.

Can these results (or their "equivalent" modulo GCH where $2^{\alpha}$ replaces ${\alpha}^+$) be proven in ZFC by taking advantage of the structure of $No(\lambda)$?