Show combinatorially that $S$ freely generates $\langle S\rangle$ Let $S = \{b^nab^{-n}: n\in\mathbb{N}\}$, $S$ freely generates $\langle S\rangle$.
The definition I have of freely generates is: $F$ is freely generated by a subset $S\subset F$ if for any group $\Gamma$ and any map $\phi:S\to \Gamma$ we can uniquely extend this to a homomorphism $\varphi:F\to S$.
I think this is essentially trivial but I'm not too comfortable with using this definition yet so would like to check. If $\phi(b^nab^{-n}) = g_n\in\Gamma$ I would just extend this multiplicatively to a homomorphism on $\langle S\rangle$ and then conclude immediately. I don't really see that there's anything at all that needs to be done here?
I'm sure I'm over simplifying it so any pointers would be excellent!

Hint: "Prove combinatorially" or "show combinatorially" almost certainly means something like this. 
Denote $c_n = b^n a b^{-n}$ and $C = \{c_n \,|\, n \in \mathbb{N}\}$. 
Given a word $w$ in the elements of $C$, let $\bar w$ denote the element of the free group $\langle a,b \rangle$ obtained by multiplying out the letters of the word $w$. For example, if $w = c_1 c_2$ then $\bar w = (bab^{-1})(b^2ab^{-2}) = babab^{-2}$. 
Prove that if $w$ is a nontrivial reduced word in the elements of $C$ then $\bar w$ is a nontrivial element of the group $\langle a,b \rangle$.
