Let $X\subset\mathbb{P}^{d}$ be a rational normal curve. After a change of coordinates, it is the image of the map:
$\nu:\mathbb{P}^{1}\rightarrow\mathbb{P}^{d}, (a_{0}:a_{1})\mapsto (a_{0}^{d}:a_{0}^{d-1}a_{1}:...:a_{1}^{d})$.
In Joe Harris's algebraic geometry book, it is said that $d+1$ distinct points of $X$ are linearly independent because the Van der Monde determinant vanishes only if two of its rows coincide. Now, let $p_{0}=\nu(a_{0,0}:a_{0,1}),...,p_{d}=\nu(a_{d,0}:a_{d,1})$ be $d+1$ distinct points of $X$. The matrix $$ \left( \begin{array}{ccc} a_{0,0}^{d} & a_{0,0}^{d-1}a_{0,1} &... & a_{0,1}^{d} \\ a_{1,0}^{d} & a_{1,0}^{d-1}a_{1,1} &... & a_{1,1}^{d} \\ \vdots & \vdots &\ddots & \vdots \\ a_{d,0}^{d} & a_{d,0}^{d-1}a_{d,1} &... & a_{d,1}^{d}\end{array} \right) $$ would be a Van der Monde matrix if $a_{0,0},...,a_{d,0}=1$. How do we know there exists a change of coordinates such that $a_{0,0},...,a_{d,0}=1$?
There are two cases to consider: one of the $a_{i,0}$ is zero, or none of them are. If none of them are, you can factor out of every line a factor $a_{i,0}^d$, and you get a Van der Monde matrix with coefficients $\frac{a_{i,1}}{a_{i,0}}$, which are all distinct. Otherwise, one of the $a_{i,0}$ is zero, without loss of generality you may assume it's the last one, then you have that the last line of your matrix is (proportional to) $$(0~0~\cdots~0~1)$$ and you develop the determinant using this line. What remains is a determinant as was discussed earlier, which is nonzero.