Surcomplex numbers and the largest algebraically closed field

Let $F$ be an algebraically closed field of characteristic zero.

Let $\kappa$ be the cardinal of a transcendance basis of $F$ over its prime subfield $\mathbb{Q}$.

Since the ${\omega_0}^{{\omega_0}^{\alpha}}$, $\alpha < \kappa$ satisfy the relation $\forall \alpha < \beta < \kappa, \forall n \in \mathbb{N}, n.({\omega_0}^{{\omega_0}^{\alpha}})^n < {\omega_0}^{{\omega_0}^{\beta}}$, they form an algebraically independant familly over $\mathbb{Q} \subset No$.

So $No[i]$, being algebraically closed, contains an isomorphic copy of $F$ as the relative algebraic closure of $\mathbb{Q}(({\omega_0}^{{\omega_0}^{\alpha}})_{\alpha < \kappa})$.

This is also true in NBG with global choice if $F$ is an algebraically closed Field of characteristic zero, and to see this one only needs to repeat the same argument with $Ord$ instead of $\kappa$.

Note that being a universal algebraically closed Field of characteristic zero caracterises $No[i]$ as a field whereas being a universal real closed Field does not characterise $No$. The theory of algebraically closed fields of a given characteristic is indeed more stable than that of real closed fields.