Brenti's total-positivity conjecture for the clean Eulerian triangle

Let $\bm{A}=\left(\left\langle\begin{matrix}n\\k\end{matrix}\right\rangle^{\rm clean}\right)_{n,k\geq 0}$ be the clean Eulerian triangle, where $\left\langle\begin{matrix}n\\k\end{matrix}\right\rangle^{\rm clean}$ counts permutations of $[n+1]$ with $k$ excedances (equivalently, $k$ descents), and satisfies

$$\left\langle\begin{matrix}n\\k\end{matrix}\right\rangle^{\rm clean}=(n-k+1)\left\langle\begin{matrix}n-1\\k-1\end{matrix}\right\rangle^{\rm clean}+(k+1)\left\langle\begin{matrix}n-1\\k\end{matrix}\right\rangle^{\rm clean}$$

for $n\geq 1$, with $\left\langle\begin{matrix}0\\k\end{matrix}\right\rangle^{\rm clean}=\delta_{k0}$. Brenti's conjecture. The clean Eulerian triangle $\bm{A}$ is totally positive. The conjecture is presented as an open problem; the source states that no proof was known there.