Let $g\in\operatorname{SL}\_n(\mathbb Z)$ such there exists $v\in\mathbb Q^n$
such that $v, gv, \dotsc, g^{n−1}v$ is a $\mathbb Q$-base of $\mathbb Q^n$ and there exists a $\mathbb Z$-base $w\_1, \dotsc, w\_n$ of $\mathbb Z^n$
such that, for every $1 \le k \le n$,
$$
\operatorname{span}\_\mathbb Q \{w\_i \mathrel| 1 \le i \le k\} = \operatorname{span}\_\mathbb Q \{g^{i−1}v \mathrel| 1 \le i \le k\}.
$$
Why is $g$ conjugate in $\operatorname{SL}\_n(\mathbb Z)$ to a matrix which belongs to the Bruhat cell $B\_\sigma$ corresponding to the permutation $\sigma \mathrel{:=} (1\ 2\ \dotsc\ n)$? 



Relation to the Bruhat cell
This is easily seen when working with matrices, so we'll work with matrices throughout. Thus, $B$ is identified with the set of upper triangular matrices, and every element $x\in \mathrm{SL}\_n(\mathbb Q)$ is idenitfied with the matrix representing it in the standard basis. In particular $\sigma$ is the permutation matrix with $0$'s on the diagonal, $1$'s in the sub diagonal and $1$ in the upper-right corner. Given a basis $\mathcal{B}$ and a matrix $x$, I write $[x]\_{\mathcal{B}}$ for the matrix representing $x$ in $\mathcal{B}$.


Write, using the OP's notation, $\mathcal{W}=\lbrace w\_1,\ldots, w\_n\rbrace$ and $\mathcal{V}=\lbrace v,gv,\ldots g^{n-1}v\rbrace$. First note that, by assumption, $[g]\_{\mathcal{W}}$ is $\mathrm{GL}\_n(\mathbb Z)$-conjugate to $g$, so, up to maybe changing $w\_1\mapsto -w\_1$, we may assume that they are $\mathrm{SL}\_n(\mathbb Z)$-conjugate.


Secondly, your assumption about the bases $\mathcal{V}$ and $\mathcal{W}$ implies that the basis-transformation matrix between them is upper triangular, and hence there exists $b\in B$ such that $[g]\_{\mathcal{V}}=b[g]\_{\mathcal{W}}b^{-1}$. 


Finally, we can write $[g]\_{\mathcal{V}}$ (more or less) explicitly: If we write $g^n=a\_0v+a\_1gv+\cdots +a\_{n-1}g^{n-1}v$, then $[g]\_{\mathcal{V}}$ is just the companion matrix with $0$ on the diagonal, $1$ on the subdiagonal and $(a\_0,\ldots,a\_{n-1})^t$ in the rightmost column. It is easy to that there exists a matrix $b\_1\in B$ such that $[g]\_{\mathcal{V}}b\_1=\sigma$. To do so, just note that you can pass from $[g]\_{\mathcal{V}}$ to $\sigma$ using column operations, and take notice regarding which operations precisely you need to perform.


From the last two paragraphs, we have that 
$$[g]\_{\mathcal{W}}=b^{-1}[g]\_{\mathcal{V}}b=b^{-1}\sigma b\_1^{-1}b\in B\_\sigma. $$
Since $[g]\_{\mathcal{W}}$ is $\mathrm{SL}\_n(\mathbb Z)$-conjugate to $g$, we are done.


