Cayley graph is a tree iff group is free I am looking at this proof of this claim that the cayley graph is a tree iff g is a free group with generating set S.
For the direction '$\implies $' I see that they have assumed that there are two paths from $v_1$ to $v_p$ implied by the fact that there is a cycle based at $v_1$ 
but i don't really understand how this assumption can be made. 
i know there exists one path following the vertices $\{v_1,v_2,...v_p\}$ but where is the second path, is it the edge going from $v_1$ to $v_p$ as they are adjacent. I thought edges in the cayley graph were directed which would imply this isn't a valid path.
any help on this would be great thank you!

 A: In the Cayley graph, each of the edges corresponds to multiplication with one of the generators -- in one direction. In the other direction, the edge corresponds to multiplication with the inverse of that generator.
Thus, every edge can be traversed in either direction -- that's what the $\epsilon_i=\pm1$ exponents in the proof you show are for.
In other words, if you have a (simple) cycle in the graph, you can go around the cycle and keep track of what you're multiplying with -- and since by definition of cycle you end up where you started -- say, at element $b$ -- you have
$$ ba_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_k^{\epsilon_k} = b $$
Canceling $b$ we get
$$ a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_k^{\epsilon_k} = e $$
The left-hand side is not forced to be $e$ by the group axioms (this would only be the case if we had $a_i=a_{i+1}$ and $\epsilon_i=-\epsilon_{i+1}$ for some $i$), and therefore the latter relation is impossible in a free group.
A: A graph is only called a tree when it is acyclic as an undirected graph, so the directions of the edges are irrelevant when considering whether a graph is a tree.  An acyclic directed graph need not be a tree, as you illustrated with your worries about the proof; a cycle may exist that is not a directed cycle.
A: Note that a graph is not tree if and onlt if it includes a cyle. But that means that a word $a_1a_2..a_n=1$. Hence the group is not free. When a group is not free then it must have none trivial relation satisfied by generaters with $a_1...a_n=1$. Hence, we have a cyle in the corresponding graph.
