$\alpha_\ell= \sigma^{p_{i_\ell}}$ as a word in $\{ \sigma^{\pm 1}, \tau^{\pm 1} \}$. Hence, $v_\ell$ is again a non-empty word ending with $\sigma^{\pm 1}$.

Case (iii): Finally, suppose the last letter of $u_\ell$ is $\sigma^{\pm 1}$. If $u_\ell$ is not a power of $\sigma$, then $u_\ell= \cdots \tau^{q} \sigma^{p}$ for some $q$ and $p$. So, $u_\ell \alpha_\ell= \cdots \tau^q \sigma^{p+p_{i_\ell}}$. Note that we must have $|p| < M$. For otherwise, the length of $u_\ell$ as a word in $\hat{\calG} \cup \hat{\calG}^{-1}$, and thus the length of $w$ as a word in $\hat{\calG} \cup \hat{\calG}^{-1}$, is at least $M$; this contradicts with the assumption on the length of $w$. Now, $|p|<M$ and $p_{i_\ell}>M$ together imply $p+p_{i_\ell} \neq 0$. Thus, the last letter of $v_\ell$ must be $\sigma^{\pm 1}$. If $u_\ell$ is a power of $\sigma$, say $u_\ell= \sigma^p$, then $$v_\ell=\begin{cases}
\sigma^{p+p_{i_\ell}} & \text{if $\ell=1$} \\
\tau^{-p_{i_{\ell-1}}} \sigma^{p+p_{i_\ell}} & \text{if $\ell \neq 1$ and $\varepsilon_{\ell-1}=+1$} \\
\sigma^{-p_{i_{\ell-1}}+p+p_{i_\ell}} & \text{if $\ell \neq 1$ and $\varepsilon_{\ell-1}=-1$}
\end{cases}$$ In the first two subcases, $v_\ell$ ends with a nontrivial power of $\sigma$, because $|p|<M$ and $p_{i_\ell} >M$ imply $p+p_{i_\ell} \neq 0$. In the third subcase, we must have $i_\ell \neq i_{\ell-1}$, and hence $|p_{i_\ell}-p_{i_{\ell-1}}|>M$. It now follows 