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.
