The surreal version of $e$ For a sequence $(x_{\alpha})$ of surreal numbers indexed by the set of all ordinal numbers, we say that $\lim x_{\alpha}=l$ ($l$ is a surreal number) if for each surreal $\epsilon>0$, there exists an ordinal $\beta$ such that $|x_{\alpha}-l|<\epsilon$ for each ordinal $\alpha>\beta$.
Consider the sequence $x_{\alpha}=(1+1/\alpha)^\alpha$, where for each surreal $x,y$, $x^y$ is defined by $\exp(y\log x)$.
Q1: What is the normal form of $x_{\omega}$? Is its real part equals to $e$?
Q2: What is $\lim x_{\alpha}$?
 A: $\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum à la Hahn series $\sum \limits_{n \in \mathbb{N}} \frac{(-1)^n\varepsilon^{n+1}}{n+1}$, and $\log(1+\varepsilon) - \varepsilon$ is negligeable with respect to $\varepsilon$. So the proof for real numbers can be applied here to show that the sequence converges to $\ee$. 
As for $x_{\omega}$, this is $\exp(1 - \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ where $\exp(- \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ is infinitesimally close to $1$ because $a:= -\frac{1}{2\omega} + \frac{1}{3\omega^2} - ...$ is an infinitesimal. So the real part of $x_{\omega}$ is $\ee$.

For now I don't see what its normal form because some combinatorial cleverness seems to be required in unfolding $\sum \limits_{n \in \mathbb{N}} \frac{a^n}{n!}$. I'll edit this answer if I find something. In the meantime you can try to find it too: we know that exponents of $\omega$ in the $a^n$ and $\exp(a)$ will be negative integers. It might be easier to compute the exponential sum directly if you can find relations between the coeffifients $q_{n,k}$ of $\omega^{-k}$ in $a^n$ for different values of $n$. 
