First some definitions (which come from Geometric Group Theory by Clara Loh)
Let $X=(V,E)$ be a graph.
Let $n \in \Bbb{N} \cup \{\infty \}$. A path in $X$ of length $n$ is a sequence $v_0,....,v_n$ of different vertices $v_0,...,v_n \in V$ with the property that $\{v_j,v_{j+1}\} \in E$ holds for all $j \in \{0,...,n-1\}$
Let $n > 2$. A cycle in $X$ of length $n$ is a path $v_0,...,v_{n-1}$ in $X$ with $\{v_{n-1},v_n\} \in E$.
The second definition is a source of confusion for me at the moment. What is $v_n$ suppose to be? Why are we stipulating that $\{v_{n-1},v_n\} \in E$? Shouldn't we be stipulating that $\{v_0,v_{n-1}\} \in E$?
A tree is a connected graph that does not contain any cycles.
Here is a theorem Loh proves in her book:
Let $F$ be a free group, freely generated by $S \subseteq F$. Then the corresponding Cayley graph $Cay(F,S)$ is a tree
Here is Loh's proof:
First, what is $s_0$ and why does $s_0 ... s_{n-1}$ being reduced follow from the elements $g_0,...,g_{n-1}$ being distinct? Second, why does $s_n ... s_1 = e$ contradict $s_0 ... s_{n-1}$ being reduced?
The only occurrence of $s_0$ is a typo for $s_1$.
If the word where not reduced, we'd have $g_{j}g_{j+1}^{-1}=s_{j+1}^{-1}=s_{j+2}=g_{j+2}g_{j+1}^{-1}$ somewhere in it, hence $g_j=g_{j+2}$.
As each $s_j$ is a generator or its inverse, the fact that $s_1\cdots s_{n-1}$ is reduced means that also the reversed word $s_{n-1}\cdots s_1$ is reduced. Then when we prepend another generator or its inverse $s_n$, at most one cancellation can occur, i.e., either $s_ns_{n-1}\cdot s_1$ is already reduced of length $n$, or $s_n=s_{n-1}^{-1}$ and after reduction we are left with a reduced word of length $n-2$. At any rate, from $n\ge 3$ we see that we cannot end up with the empty word.