TITLE: Tighter Caratheodory on the moment curve?
QUESTION [12 upvotes]: The moment curve is the set of points of the form 
$$(t,t^2,t^3,...,t^n) \in R^n$$
Let $M$ be the portion of the moment curve where $t\in [0,1]$, and let $\overline{M}$ be the convex hull of $M$.
Caratheodory's theorem tells us that any point in $\overline{M}$ can be expressed as a convex combination of $n+1$ points in $M$.  However, when $n=1$ and $n=2$, we can get by with only $n$ points instead of $n+1$ points.
Question: Is this true for all $n$?  I.e., can we express any point in $\overline{M}$ as a convex combination of at most $n$ points from $M$?
Edit: In the answer below, eins6180 points out that $n$ points always suffice, answering my question.  However, the reference he suggested actually mentions something much stronger.  The moment curve is a convex curve, that is, a curve in which no $n + 1$ points lie in a single affine hyperplane.  Barany & Karasev mention (through reference) that convex curves only require
$$\left\lfloor \frac{n+2}{2} \right\rfloor$$
points to represent any point in their convex hull.  The moment curve example is discussed a little bit in this paper by Holmsen and Roldan-Pensado.
I think a dimension-counting argument shows that this bound is tight.

REPLY [13 votes]: The answer is yes for all dimensions.
An old theorem by Fenchel states that for a compact set $K$ in $\mathbb{R}^n$ every point in the convex hull can be either written as a convex combination of at most $n$ points or $K$ can be separated by a hyperplane. Since $M$ is path-connected, such a separation is not possible.
But there are also more modern theorems of this kind available. A good starting point is recent work by Barany & Karasev. Their work also implies an affirmative answer to your question.