If I consider a word $p_0^{k_0}\cdots p_{n-1}^{k_{n-1}}$ of $N,$ with $k_0,\dots,k_{n-1}\in\mathbb{Z}$ I could rewrite it $p_0^{l_0}\cdots p_{n-1}^{l_{n-1}}$ with $l_0,\dots,l_{n-1}\in\{0,\dots,n-1\}.$ I can simplify it as $q^jp^{l_j}qp^{l_{j+1}}q\cdots p^{l_{n-1}}q.$ where $j$ is the smallest non negative integer such that $p_j=q^jp^{l_j}q^{-j}\neq0.$ I don't see how it is possible to write this word in terms of $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}.$

I tried to prove this statement but I am stucked. I set for $k\in\{0,\dots,n-1\},p_k=q^kpq^{-k}=q^kpq^{n-k}.$ To prove that $N$ is free on $p,p^q,p^{q^2},\ldots,p^{q^{n-1}},$ is it sufficient to prove that any word $p_0^{k_0}\cdots p_{n-1}^{k_{n-1}}$ where $k_0,\dots,k_{n-1}\in\mathbb{Z}$ can be written in words in $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}?$

It's ok for the index of $N.$ My bad

Is the induction of a faithful representation over $\mathbb{Z}$ always faithful?

I have never heard of Kurosh theorem before. I read an article on Wikipedia and according to it, we have that normal closure of p say $N:=\langle q^kpq^{−k}|k∈{0,…,n−1}\rangle$ is the a product $F(X)*(*_{i\in I}A_i)*(*_{j\in J}B_j)$ where $X$ is a subset of $H$, $(A_i)_{i\in I}$ and $(B_j)_{j\in I}$ are respectively families of subgroups of $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ But I don't see why we can get rid of the last two components to say $N$ is free and I don't see why $N$ has index n

where can I find a detailed proof of the fact that a free group has a faithful representation over $\mathbb{Z}$ of degree $2?$

yes I did. In that article, Brunner & Sidki want to prove that $GL_n(\mathbb{Z})$ is an automaton group. But I think it's not the case. Anyway I don't see any link with the fact that $BS(n,n)$ is an automaton group, $F_{|n|}\rtimes\mathbb{Z},$ and the fact $GL_n(\mathbb{Z})$ is an automaton group. The only link (which is false) I see is even if $BS(n,n)$ is a subgroup of $GL_n(\mathbb{Z})$ and $GL_n(\mathbb{Z})$ is an automaton group, it does not make $BS(n,n)$ an automaton group.

I never heard of "standard Bass-Serre tree" of a group? What is it?

In fact I meant both statements. sorry for the confusion

yes I'm also asking why (1) holds

Thank you for your answer. Why does this property make $BS(n,n)$ an automaton group?