= Solution
Apply a <projective linear transformation> sending the first $n+1$ points to the <coordinate points>
$$
e_0=[1:0:\cdots:0],\ldots,e_n=[0:\cdots:0:1].
$$
The next point has every <homogeneous coordinate> nonzero by <general linear position>, so a diagonal projective transformation fixing the $e_i$ sends it to
$$
e=[1:\cdots:1].
$$
Write the last point as $a=[a_0:\cdots:a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the <hyperplane> $x_i-x_j=0$.
Put $t_i=a_i^{-1}$ and define the degree-$n$ <homogeneous polynomials>
$$
P_i(S,T)=\prod_{j\ne i}(S-t_jT),\qquad 0\leq i\leq n.
$$
The associated map $\mathbb P^1\to\mathbb P^n$ is a <rational normal curve>. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, because
$$
P_i(0,1)=(-1)^n\prod_{j\ne i}t_j
$$
is one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.
For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source $\mathbb P^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is therefore
$$
Q_i(S,T)=c_i\prod_{j\ne i}(S-s_jT).
$$
The value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on $\mathbb P^1$. This merely reparametrizes the curve, proving uniqueness.
For $n=3$, choose the <monomial basis> $S^3,S^2T,ST^2,T^3$. The resulting curve is the <twisted cubic>
$$
[S:T]\longmapsto[S^3:S^2T:ST^2:T^3].
$$
Its <homogeneous ideal> is generated by the three quadratic <matrix minors>
$$
\boxed{x_0x_2-x_1^2,\qquad x_0x_3-x_1x_2,\qquad x_1x_3-x_2^2},
$$
the $2\times2$ minors of
$$
\begin{pmatrix}x_0&x_1&x_2\\x_1&x_2&x_3\end{pmatrix}.
$$
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