Order of the product of consecutive adjacent transpositions Given the generators $\sigma_1,\dots,\sigma_{n-1}$ that satisfy $\sigma _{i}^{2}=1$, $\sigma _{i}\sigma _{j}=\sigma _{j}\sigma _{i}$ for $|i-j|>1$, and $(\sigma _{i}\sigma _{i+1})^{3}=1$, prove that $(\sigma_1\sigma_2\cdots\sigma_{n})^{n+1} = 1$.
I am looking for a direct derivation of the order of the $n$-cycle from the properties of transpositions.
For example, for $n=1$ and $n=2$ the statement is found among the basic properties: $\sigma_1^2=1$ and $(\sigma_1\sigma_2)^3=1$.
For $n=3$, we can derive $\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}$ from the basic properties and then use it to show that
$$
(\sigma_1\sigma_2\sigma_3)^2=\sigma_2\sigma_1\sigma_3\sigma_2
$$
and then using $\sigma_i^2=1$ and commutativity of $\sigma_1$ and $\sigma_3$ we get
$$
(\sigma_1\sigma_2\sigma_3)^4=(\sigma_2\sigma_1\sigma_3\sigma_2)^2=\sigma_2\sigma_1\sigma_3\sigma_1\sigma_3\sigma_2=
\sigma_2\sigma_1\sigma_1\sigma_3\sigma_3\sigma_2=1.
$$
How to prove this for an arbitrary $n$?
 A: Let's present the proof for the case $n=4$ in the form that will be amenable to generalization.  Introducing a bit of notation,
$$
(i:j)=\sigma_i\sigma_{i+1}\cdots\sigma_j,
$$
we can rewrite the desired identity as
$$
(1:4)^5 = 1
$$
We will prove this by computing the powers of $(1:4)$ consecutively as follows:
$$
\begin{align}
(1:4)&=(1:3)\,\sigma_4\\
(1:4)^2&=(1:3)^2\sigma_4\sigma_3\\
(1:4)^3&=(1:3)^3\sigma_4\sigma_3\sigma_2\\
(1:4)^4&=(1:3)^4\sigma_4\sigma_3\sigma_2\sigma_1=(1:4)^{-1}\\
\end{align}
$$
Note that in order to arrive at the last identity we used $(1:3)^4=1$ that was demonstrated in the formulation of the question.
Extending $(i:j)$ notation to the case $i>j$ as
$$
(i:j)=\sigma_i\sigma_{i-1}\cdots\sigma_j,
$$
we can generalize the equations for the powers of $(1:4)$ above to
\begin{equation}\tag{1}\label{rec}
(1:n)^k = (1:n-1)^k(n:n-k+1)
\end{equation}
which we will prove by induction. The base case, $k=1$ follows directly from the definitions:
$$
(1:n)=\sigma_1\cdots\sigma_{n-1}\sigma_n = (1:n-1)(n:n)
$$
To prove the induction step, we assume that \eqref{rec} holds and will derive
\begin{equation}\tag{2}\label{rec2}
(1:n)^{k+1} = (1:n-1)^{k+1}(n:n-k)
\end{equation}
Indeed, starting with
$$
(1:n)^{k+1} = (1:n)^k(1:n)
$$
and using \eqref{rec}, we find
\begin{equation}\tag{3}\label{rec3}
(1:n)^{k+1}=(1:n-1)^k(n:n-k+1)\,(1:n)=(1:n-1)^k(n:m)\,(1:n),
\end{equation}
where $m=n-k+1$.
Since $\sigma_i$ commutes with $\sigma_j$ whenever $|i-j|>1$, $(1:m-2)$ part of $(1:n)$ commutes with $(n:m)$ and we can bring it to the left and rewrite \eqref{rec3} as
\begin{equation}\tag{4}\label{rec4}
(1:n)^{k+1}=(1:n-1)^k(1:m-2)\,(n:m)\,(m-1:n).
\end{equation}
We will now focus on the last two multiplicands
$$
(n:m)\,(m-1:n)=(n:m+1)\sigma_{m}\sigma_{m-1}\sigma_{m}(m+1:n)
$$
where the product $\sigma_{m}\sigma_{m-1}\sigma_{m}$ in the middle can be rewritten as $\sigma_{m-1}\sigma_{m}\sigma_{m-1}$. Using commutativity of $\sigma_{m-1}$ with non-adjacent ranges
$$
\begin{align}
(n:m)\,(m-1:n)&=(n:m+1)\sigma_{m-1}\sigma_{m}\sigma_{m-1}(m+1:n)\\
              &=\sigma_{m-1}(n:m+1)\sigma_{m}(m+1:n)\,\sigma_{m-1}\\
              &=\sigma_{m-1}(n:m+1)\,(m:n)\,\sigma_{m-1}
\end{align}
$$
Note that in the result the product sandwiched between the sigmas is the same as the original but with $m$ incremented by $1$.  Therefore we can repeat the same transformation for increasing $m$ until $m=n-1$ and we get
$$
(n:n)\,(n-1:n)=\sigma_{n}\sigma_{n-1}\sigma_{n}=\sigma_{n-1}\sigma_{n}\sigma_{n-1}
$$
Taking into account the sigmas that appear on both sides in every step we conclude that
$$
(n:m)\,(m+1:n)=(m-1:n-1)(n:m-1)
$$
Finally, substituting the last identity in \eqref{rec4}
$$
\begin{align}
(1:n)^{k+1}&=(1:n-1)^k(1:m-2)\,(n:m)\,(m-1:n)\\
           &=(1:n-1)^k(1:m-2)\,(m-1:n-1)\,(n:m-1)\\
           &=(1:n-1)^{k+1}\,(n:m-1)
\end{align}
$$
we obtain \eqref{rec2}.  Q.E.D.
