The set up is as stated above. We have a projective curve $X$ of degree n embedded in $\mathbb{P^n}$, which is not contained in any hyperplane. We claim that it is therefore rational.
The way I have been told to tackle it, is to project the curve from a point down to $\mathbb{P^{n-1}}$, where we will get a curve of degree n-1. Proceeding in this way we reduce to the case of a conic in $\mathbb{P^2}$, which is rational.
However, what I don't see is why the image of this map gives us a curve of degree n-1. I can sort of intuitively see it, but I'm having a hard time coming up with a rigorous argument.
I think the trick is to project from a general point $p$ lying on $X$. Let $\pi : \mathbf P^n \dashrightarrow \mathbf P^{n-1}$ be the projection. Then any hyperplane $H$ in $\mathbf P^{n-1}$ has the form $\pi (H')$ for some hyperplane $H' \subset \mathbf P^n$ passing through $p$. Now a general $H'$ intersects $X$ in precisely $n$ points, one of which is $p$. The other $n-1$ intersections points will give the intersection points of $H$ with $\pi(X)$. This shows that $\pi(X)$ has degree $n-1$.