I know this is a standard result and I'm looking at the proof here (Theorem 2.0.1)
I have one quick question about this proof.
The proof uses induction on $n$ by considering any permutation $p\in S_n$ and showing that there exists a product of adjacent transpositions $q$ s.t. $q\circ p(n)=n$. Then $q\circ p$ can be viewed as a permutation in $S_{n-1}$.
I understand what $q$ is and that $q\circ p\in S_{n-1}$ and thus $q\circ p$ can be expressed as a product of adjacent transpositions by induction hypothesis. But why does that mean $p$ can be expressed as a product of adjacent transpositions?
You have that $q$ is a composition of adjacent transpositions, and so is $q\circ p$. But $p=q^{-1}\circ(q\circ p)$, and so $P$ is also a composition of adjoint transpositions.