  Let $P$ be a set of $n$ points in $\mathbb{R}^d$, in general position. We
remove all of them one by one, in each step erasing one vertex of the convex
hull of the current remaining set. Let $g_d(P)$ denote the number of different
removal orders we can attain while erasing all points of $P$ this way, and let
$g_d(n)$ be the \emph{minimum} of $g_d(P)$ over all $n$-element point sets
$P\subset \mathbb{R}^d$. Dumitrescu and T\'oth showed that
$g_d(n)=O((d+1)^{(d+1)^2n})$.
  We substantially improve their bound, by proving that $g_d(n)=
O((d+d\ln{(d)})^{(2+\frac{(d-1)}{\lfloor d\ln{d}\rfloor})n})$. It follows that,
for any $\epsilon>0$, there exist sufficiently high dimensional point sets
$P\subset \mathbb{R}^d$ with $g_d(P)\leq O(d^{(2+\epsilon)n})$. This almost
closes the gap between the upper bound and the best-known lower bound
$(d+1)^n$.
