Powers of Zariski-dense subset span symmetric power

I wonder if something like the following is true: let $V$ be a finite-dimensional vector space over a field of characteristic zero and $S \subset V$ a Zariski-dense subset. Does $\{ v^d \ | \ v \in S \}$ span $\text{Sym}^d(V)$? If $S$ is a full lattice this is classical, but perhaps the proof of that result could be simplified by arguing in this way?

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Oops, I already figured it out. If $\{ v^d \ | \ v \in S \}$ is contained in some hyperplane $H \subset \text{Sym}^d(V)$, write $H = \{ L = 0 \}$ where $L \in \text{Sym}^d(V)^* \cong \text{Sym}^d(V^*)$. Then thinking of $L$ as a degree $d$ form on $V$, we see that $L$ vanishes on $S$ and hence $V$ by density, so $L$ must be the zero form.