This is an exercise from Beauville's book on complex surfaces. The claim is that for any irreducible surface $S \subseteq \mathbb{P}^{n}$ of degree $\le n-2$, $S$ is contained in a hyperplane.
In my first approach I tried contradiction. I assumed that $S$ was contained in $\textit{no}$ hyperplane. Using the exact sequence (where $E$ is a hyperplane section of $S$)
$$0 \rightarrow \mathscr{O}_{S} \rightarrow \mathscr{O}_{S}(E) \rightarrow \mathscr{O}_{E}(E) \rightarrow 0, $$
and Riemann-Roch, one can show that
$$n-1 \le h^{0}(S,\mathscr{O}_{E}(E)) \le E^{2}-1+h^{1}(S,\mathscr{O}_{E}(E)).$$
My only snag is that I am not sure how to compute $h^{1}(S,\mathscr{O}_{E}(E))$. In addition, my professor said I should look for a more intuitive and geometric solution.
$\textbf{Question:}$ Am I on the right track with this proof, or is there a simpler geometric solution which someone could direct me toward?
I was stuck on the same problem just now, which is how I came upon here, but I've figured it out. Even though it seems that I'm late to the party, I'll post it here for anyone else who might happen to see this.
The idea is that $n$ points in $\mathbb P^n$ determine a hyperplane. Assume that $S$ is not contained in a hyperplane. Then there are $n$ distinct points on $S$ which determine a hyperplane $H$. Through $n-1$ of these points lies some other hyperplane $H'$. Then on $S$, we have $n-2\geqslant \deg S = H^2 = H.H' \geqslant n-1$, a contradiction. Thus $S$ is contained in a hyperplane.