$PGL(n + 1)$ acts on sets of $n + 2$ points in $\mathbb{P}^n$ transitively: proof without determinants? 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?
 A: 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 $j=0,\dots,n$.
