Given a reduced expression for the longest word $w_0$ in the Weyl group of $\mathfrak{g}=\mathfrak{n}^+\oplus\mathfrak{h}\oplus{n}^-$, one obtains a convex ordering on the set of positive roots, $\beta_1<\ldots<\beta_N$. Using the braid group action we get (divided powers of) root vectors $E_{\beta_i}^{(n_i)}$ in $U_q(\mathfrak{n}^+)$ and the set of elements of the form $$E(\underline{n})=E_{\beta_N}^{(n_N)}\cdots E_{\beta_1}^{(n_1)},\;\;\;\underline{n}=(n_N,\ldots,n_1)\in\mathbb{Z}^N_{\geq0}$$ forms a basis for $U_q(\mathfrak{n}^+)$.

I'm wondering if anyone can point me to an elementary proof of the following fact (the shorter the better):

$$\overline{E(\underline{n})}=\sum_{\underline{m}}\lambda_{\underline{n}\underline{m}}E(\underline{m})$$ where $\lambda_{\underline{n}\underline{n}}=1$ and $\lambda_{\underline{n}\underline{m}}=0$ if $\underline{m}>\underline{n}$ with respect to some total ordering. (I think the lexicographic ordering on $\mathbb{Z}_{\geq0}^N$ or maybe the opposite ordering works, but if not I'd like to know what does).

By elementary, I mean that I want to avoid the geometry of quiver varieties. Also, I know that Leclerc has a proof of this fact in his paper on quantum shuffles, but I'm hoping for something less involved.