In this famous MO question, a beautiful proof is given of the fact $V\cong V^\ast\iff V$ is finite dimensional.
I'm trying to go through it and I'm having some trouble. First of all, I know the in the infinite dimensional case, $|V^\ast|=|k|^{\dim V}$ and $|V|=\max (|k|,\dim V)$. Now, the assumption in the first claim says $\dim V$ is at least $|k|$. If this means a loose inequality, then I don't understand how we can conclude $\dim V< \dim V^\ast$, since the only result on cardinal exponentiation I know of which gives a strict inequality is:
If $\kappa$ is infinite with $\mathrm{cf}(\kappa)\leq \lambda$ then $\kappa ^\lambda >\kappa$.
If we do assume a strict inequality, then I still don't know why our inequality follows, but even worse, I don't understand why knowing that each field contains a coutnable subfield finishes the proof. So I'm pretty sure we only need to assume a loose inequality, but I just don't understand the cardinal arithmetic.
The second part of my question is about which parts of the proof fail when generalizing to free non-finitely generated modules over commutative rings. It seems the problem can only arise in proposition 2, since the first one involves cardinality games alone, and the last fact seems to be true for most algebraic structures (having a countable substructure).
For any $V$, we have $|V^*|=|k|^{\dim V}\geq 2^{\dim V}>\dim V$. The point of the assumption $\dim V\geq |k|$ is that it implies (via $|V|=\max(|k|,\dim V)$) that $\dim V=|V|$, so we get $|V^*|>|V|$, which then clearly implies $\dim V^*>\dim V$.
The case of arbitrary commutative rings actually follows more or less immediately from the field case. If $A$ is a (nonzero) commutative ring, choose any maximal ideal $m$, with residue field $A/m=k$. If $F=A^{\oplus S}$ is a free module with basis $S$, then $F^*$ is the infinite product $A^S$. Note that $F/mF$ can naturally be identified with $k^{\oplus S}$, and $F^*/mF^*$ has a natural surjection to $k^S$, and in particular, $\dim_k F^*/mF^*\geq\dim k^S$. If $F$ is isomorphic to $F^*$, then $F/mF$ must be isomorphic to $F^*/mF^*$ as a $k$-vector space. This means $|S|=\dim_k F^*/mF^*\geq \dim k^S$. By the field case, this implies $S$ must be finite.