I am trying to understand the following passage from Skolem’s (1922) proof of the Lowenheim-Skolem theorem by reference to the contemporary proof of Konig’s infinity lemma:

>Let $L_{1,n},L_{2,n},...,L_{e_n,n}$ be the solutions of the nth step. If we now form the sequence $L_{1,1},L_{1,2},...$ of the first solutions, we can verify without difficulty that they converge in the logical sense. For let $L_{1,n}$ be a continuation of $L_{{a_v^n}_v}$ $(n> v)$. Then, if $n'> n$, ${a_v^n}_v$ ≦ ${a_v^{n'}}_v$  . But, since the
number ${a_v^n}_v$ can only have values 1 to $e_v$ it must remain constant for
all sufficiently large $n$. Thus we can obtain as "limit" the fact that the
first-order proposition is satisfied in the domain of the entire number
sequence, q.e.d. (Skolem 1922, p. 294)


Skolem’s “solutions” are assignments of truth-values to the atomic components of a propositional formula U, and each step considers progressively more instances of U, as part of the process of constructing a model of U’s quantified counterpart F. The ordering is such that $L \leq L'$ if the first atomic component of U on which they differ gets assigned FALSE in $L$ and TRUE in $L'$.

What I *thought* was going on in Skolem’s proof was that after singling out a particular path through the tree of “solutions”, i.e., the path consisting of the first (w.r.t the given ordering) solution for each level, he just needs to show that $L_{1,n} \subset L_{1,n+1}$. This is because we need a path through the tree that assigns truth values to atomic components once and for all, so that solutions higher up in the construction don’t disagree on their assignments to instances of U already present at earlier levels. 

If that is correct, I have trouble seeing the argument for “logical convergence” in this light. As I understand it, every first solution of the nth step $L_{1,n}$ extends some (unique?) solution of the vth level. Let ${a_v^n}_v$ be the place in the ordering (of the vth level solutions) at which this solution occurs. Then, moving up to the n+1th level say, the first solution $L_{1,n+1}$ is also a continuation of some vth level solution, but one that must occur at the same place or later, i.e. further right, in the ordering of the vth level. If it occurred earlier, this would mean that fewer of the vth level instances of U were made true by the solution that $L_{1,n+1}$ extends, than by the solution that $L_{1,n}$ extends. 

First question: why is that possibility, namely, that ${a_v^n}_v$ > ${a_v^{n'}}_v$ ruled out? 
	
Next, the horizontal sequence of solutions is finitely bounded. So it is not possible that as the levels increase, the first solution for each n is a continuation of a higher and higher solution at the vth level, because at some point we would reach the last of the vth level solutions. 

Second question: why can’t the number $e_v$ increase at the same rate as n?

So for some n(v), every first solution after the n(v)th level will have the same assignment to instances at the vth level, i.e., the truth assignment up to the vth level is fixed. In the limit, as n goes to infinity, all the levels get fixed in this way. 
If this is correct, it seems like the desired path is not the sequence of first solutions at every level. What we want is the ${a_v^n}_v$, for each level v, such that $L_{{a_v^n}_v}$ is fixed by some n(v)th level. But then 

Thirdly: What role is the sequence $L_{1,1},L_{1,2},...$ playing in this proof, and does Skolem actually establish that $L_{1,n} \subset L_{1,n+1}$?