We work over $\mathbb{C}$. Let us define the rational normal curve of degree $d$ as the image of the morphism $$\nu_d:\mathbb{P}^1\to \mathbb{P}^d,\quad [x:y]\mapsto [x^d:x^{d-1}y:\ldots:xy^{d-1}:y^d].$$ Let $p\in \mathbb{P}^d$ be a point not belonging to $\nu_d(\mathbb{P}^1)$ and also not belonging to an hyperplane $H\simeq \mathbb{P}^{d-1}$ and consider the projection $\pi: \mathbb{P}^d \dashrightarrow H$. Obviously the restriciton of the projection to $\nu_d(\mathbb{P}^1)$ is a morphism.
Question: Is it true that $\pi(\nu_d(\mathbb{P}^1))$ is isomorphic to the rational normal curve of degree $d-1$? Intuitively I can consider the two Veronese embedding, and the only way to make the diagram commute I suspect is to project. However, I think this is well known -if it is true -and I would like to have a reference. Moreover, if this holds I guess the projection from a $\mathbb{P}^k$ of $\nu_d(\mathbb{P}^1)$, with $k<d$, will be the rational normal curve of degree $d-k-1$, right? Thanks in advance!
As noted in the comments, an easily accessible counterexample is that the projection of the twisted cubic is either the cuspidal or nodal cubic.
If it does happen that the projection gives an isomorphism (ie the point you project from is not on any tangent or secant line to the curve) then you will again get smooth rational curve, but it is not the rational normal curve because it has the wrong degree: given a hyperplane intersecting $\pi(C)$ transversely in $\deg \pi(C)$ points (such a thing always exists, ref), the hyperplane obtained by taking the closure of the preimage of this hyperplane under the projection map intersects $C$ transversely in $\deg \pi(C)$ points too - so $\deg \pi(C)=\deg C$.
Let me also point out that the "normal" in "rational normal curve" is referring to projective normality, not the property of being a normal scheme. Consider $[t^4:t^3u:tu^3:u^4]$, which is rational and normal and a projection of the 4-uple embedding but it is not projectively normal because the projective coordinate ring is not integrally closed: $t^2u^2=\frac{(t^3u)^2}{t^4}$ solves $x^2=t^4u^4$ despite not being in the homogeneous coordinate ring.