I'm trying to solve Exercise I.13.34 of Kunen's Set Theory, which goes as follows (paraphrased):
Let $F$ be a field with $|F| < \beth_\omega$, and $W_0$ a vector space over $F$ with $\aleph_0 \le \dim W_0 < \beth_\omega$. Recursively let $W_{n+1} = W_n^{**}$ so that $W_n$ is naturally identified with a subspace of $W_{n+1}$. Then let $W_\omega = \bigcup_n W_n$. Show that $|W_\omega| = \dim W_{\omega} = \beth_\omega$.
Some useful facts:
If $W$ is a vector space over $F$ with basis $B$, there is an obvious bijection between $W^*$ and ${}^{B}F$ (i.e. the set of functions from $F$ to $B$, denoted this way to avoid ambiguity with cardinal exponentiation). Hence $|W^*| = |F|^{\dim W}$.
Asaf Karagila showed in this answer that $|W| = \max(\dim W, |F|)$.
By the "dual basis" construction we have $\dim W^* \ge \dim W$. (There's an assertion on Wikipedia that the inequality is strict whenever $\dim W$ is infinite, but I don't immediately see how to prove that.)
One inequality is pretty easy. Using Fact 1, we get $|W^*| = |F|^{\dim W} \le |F|^{|W|}$. Now thanks to the simple fact that ${}^{\beth_n} \beth_m \subset \mathcal{P}(\mathcal{P}(\beth_m \times \beth_n))$ we have $\beth_m^{\beth_n} \le \beth_{\max(m,n)+2}$. So by induction it follows that $|W_n| < \beth_\omega$ for each $n$, and hence (using Kunen's Theorem 1.12.14) we get $|W_\omega| \le \beth_\omega$.
For the other direction, if $\dim W \ge |F|$ then Fact 3 gives us $\dim W^* \ge |F|$ and hence by Facts 1 and 2 $$\dim W^* = \max(\dim W^*, |F|) = |W^*| = |F|^{\dim W} \ge 2^{\dim W}.$$
So if $\dim W_0 \ge |F|$ then by induction we get $\dim W_n \ge \beth_{2n}$ and therefore $\dim W_\omega \ge \beth_\omega$. Since $|W_\omega| \ge \dim W_\omega$ we must have equality throughout.
But I am stuck on the case $\aleph_0 \le \dim W_0 < |F|$. Intuitively it still seems like $\dim W^*$ should be "much larger" than $\dim W$. We shouldn't really need to go through the cardinalities of the spaces themselves, but I can't see what to do. Any hints?
