It is "well known" that if $p_1, \dots, p_{n + 1}$ are points in $\mathbb{P}^{n -1}$ (over $\mathbb{C}$, say) in general position, and $q_1, \dots, q_{n + 1}$ are another set of such points, then there is a unique matrix in $G \in PGL(n - 1)$ sending the first set to the second.
Not wishing to take the author's assertion for granted, I came up with the following proof. Take $p_i$ to be the standard basis vector in $\mathbb{C}^n$ for $1 \leq i \leq n$, and $p_{n + 1} = (1, 1, \cdots, 1)$. Let $A$ be the $n \times n$ matrix constructed by taking the first $n$ of the $q$'s as column vectors, then we must have $G = A D$ for some diagonal matrix $D$, none of whose entries is $0$.
Let $d$ be the vector consisting of the diagonal elements of $D$, and $v$ be the vector consisting of the coordinates of $q_{n + 1}$. Then the requirement that $G$ send $p_{n + 1}$ to $q_{n + 1}$ amounts to solving the equation $A d = v$. We see that $A$ must be invertible, and writing $d = A^{-1} v$ and using the cofactor expansion for $A^{-1}$, the condition that each $d_i$ be nonzero can be seen to be equivalent to the condition that the points other than $p_i$ are in general position: $A^{-1} v$ is equal, up to a scalar, to the vector of the $n$ relevant matrix determinants.
This proof is fine, and perhaps optimal in some sense: each of the hypotheses gets used exactly once. But it requires an arbitrary choice at the beginning, and relies on messy formulas for matrix determinants. Can this be avoided?
I'm not quite sure how you got so involved with determinants. Do we agree that for $n+1$ points in $\mathbb P^{n-1}$ to be in general position means that any $n$ of them span $\mathbb P^{n-1}$?
First, your "arbitrary choice" at the beginning is just fine. If you show that there's a group element carrying the $n+1$ "standard" points $P_0=[1,0,\dots,0]$, $P_1=[0,1,\dots,0]$, $\dots$, $P_{n-1}=[0,0,\dots,1]$, and $P_n=[1,1,\dots,1]$ to $n+1$ arbitrary points in general position, then you get the general case by composition (using the inverse in the first case): $Q_0,\dots,Q_n\rightsquigarrow P_0,\dots,P_n \rightsquigarrow R_0,\dots,R_n$.
A more conceptual way to understand what you're doing is this. Given $Q_0,\dots,Q_n\in\mathbb P^{n-1}$ in general position, we can choose homogeneous coordinate vectors $v_0,\dots,v_n\in\mathbb C^n$ so that $v_n = \sum\limits_{j=0}^{n-1} v_j$. Namely, for any choice, by general position, we have $v_n = \sum\limits_{j=0}^{n-1}\lambda_j v_j$ for some $\lambda_j\ne 0$. Now rescale. So now you take your matrix $A$ to have columns $v_0,\dots,v_{n-1}$ and clearly $AP_j = Q_j$ for all $j=0,\dots,n$.