Prove: In a finite geometry, if a line has $n$ points, then the total number of points is $n^2 - n + 1$. The following axioms define a finite geometry:
1) Any two distinct points are a unique line.
2) If p and q are distinct lines then they is a unique point on both.
3) There are at least 3 distinct points on any line.
4) Not all the points are on the same line.
Assume there exists a line $l$ with $n$ points: $p_1,...p_n$. From 4) there exists $p'$ in the geometry and not on the line. With the pairs $(pi,p')$, I can form $n$ lines. I know that each of these lines has at least 3 points, but I don't see why they can't be one of the $n_j$. I might be missing something, but I can't seem to figure out how to use 2) and 3) to get more points.
Any help would be welcome!
The stock proof of this is the one you started with.
Let $\ell $ be any line, with points $P_1, \ldots, P_n$ (where $n$ is at least $3$). Let $Q$ be any point not on $\ell$ (axiom 4). Consider the lines $$ \ell_i (i = 1, \ldots, n) $$ where $\ell_i$ contains both $Q$ and $P_i$. These are all distinct (axiom 1), and they intersect pairwise exactly at the point $Q$ (axiom 2).
Claim: every point of the projective plane lies on one of the $\ell_i$. Suppose point $A$ does not. Then the line $QA$ ($Q$ and $A$ are distinct, because $Q$ lies on all the $\ell_i$, and $A$ lies on none of them!) is parallel to $\ell$, which is impossible by axiom 2.
Claim: for any distinct $i, j$, $\ell_i$ and $\ell_j$ have the same number of points.
Proof: let $P_k$ be different from $P_i$ and $P_j$. Then $P_k$ lies on neither $\ell_i$ not $\ell_j$ (why not? You need to verify this!).
Let the points of $\ell_i$ be $A_1 \ldots A_p$; consider the lines $m_i$ from $P_k$ through $A_i$. The line $m_i$ intersects $\ell_j$ in some point $B_i$. And this provides a bijection from the $A$s to the $Bs$.
Claim: the number of points on $\ell_1$ is the same as the number $n$ of points on $\ell$.
Reason: Consider the line $\ell_2$. It contains the points $Q$ and $P_2$, and must contain another point $S$ distinct from these, by axiom 4. $S$ is not a point of $\ell$, for then $\ell$ and $\ell_2$ would both contain $P_2$ and $S$, and hence be the same (axiom 2). Nor, for similar reasons, is $S$ a point of $\ell_1$. Now for each point $P_i$ of $\ell$, define a point $R_i$: $R_i$ the the intersection of the line $SP_i$ with $\ell_1$. This defines a bijection from $\ell$ to $\ell_1$, hence they have the same number of points.
Claim: the total number of points is $n^2 - n + 1$. For there are $n$ lines $\ell_i$, each with $n$ points, for a total of $n^2$ points. But the point $Q$ lies on all of them, which means it's been overcounted $n-1$ times. QED.