Proof of the number of nodes in a finite projective plane A finite projective plane is a hypergraph in which
1. Any two edges shere exactly one node
2. There is exactly one edge containing any given pair of nodes
3. You can choose 4 nodes such that no three are contained in a common edge
I'm trying to solve an exercise to prove that for any such object, the number of nodes is $n^2+n+1$ for some n.  I pretty much have a proof, which I will post as an answer.  However, what I'm really looking for is a cleaner proof or way of thinking about it, as my method seems way too convoluted.
 A: If we have a line $L$ containing $s$ points and a point $P$ not on $L$, then the $s$ lines joining $P$ to the points of $L$ are distinct and account for all of the lines through $P$.  Hence all points not on $L$ have the same number $s$ of lines passing through them.
Likewise, if we have a point $P$ with $t$ lines through it and a line $L$ that does not pass through $P$, then the $t$ points of intersection of the lines through $P$ with $L$ are distinct, and account for all of the points of $L$.  Hence all of the lines that do not pass through $P$ contain the same number $t$ of points.
Axiom 3 guarantees us a quadrangle.  Let's say its points are $a$, $b$, $c$, $d$.  Let the number of points on the line $\overline{cd}$ be $n+1$.  Neither $a$ nor $b$ lies on $\overline{cd}$.  Therefore both $a$ and $b$ have $n+1$ lines passing through them.  Since neither $b$ nor $c$ lies on $\overline{ad}$, the same number of lines pass through each, so $c$ also has $n+1$ lines passing through it.  Now every line of the plane avoids at least one of $a$, $b$, $c$.  Hence every line contains $n+1$ points.
Furthermore, every point lies off of at least one line, which implies that every point has $n+1$ lines passing through it.  To see that every point lies off of at least one line, observe that $\overline{ab}$ and $\overline{cd}$ have exactly one point in common - call it $e$.  Every point other than $e$ lies off of at least one of $\overline{ab}$ and $\overline{cd}$, while $e$ itself lies off of $\overline{ac}$ since $a$, $b$, $c$, $d$ is a quadrangle.
To count the points of the plane, let $P$ be a point.  Every point of the plane except for $P$ lies on exactly one of the $n+1$ lines through $P$.  Each of these lines contains $n$ points apart from $P$, for a total of $(n+1)n+1$ points.   That the number of lines in the plane is also $n^2+n+1$ can be shown similarly.
A: My idea at the moment goes as follows:
Given an arbitrary projective plane, take the (one of the) longest line(s), and call the number of nodes in this line $n+1$.  Note that $n \ge 2$.  Separate one node out as the "leader" node.  We know that there are at least two other nodes because of rule 3.  There are two possibilities for how these nodes add on, given in the picture below (for the $n=4$ case).  The leftmost node in each configuration is the "leader."

Now, in the left configuration, there must be a line from 1a to 1b, as well as from 2a to 1b.  These cannot both contain 1c, but they must cross the line going from the leader to 1c. Therefore, there must be a "2c" node.  This reduces the left configuration to a stronger version of the right configuration.
Now, for the right configuration.  There is one line going down from each $1a \dots na$ through each $1b \dots kb$.  Each of the lines $2a\to2b \dots na\to2b$ must intersect the line $1a\to1b$, but since they have all already touched, they must each meet $1a\to1b$ at a new node, so we get $n-1$ new nodes.  These must all be in different rows, so we now have a total of $n+1$ rows.  Sort these rows by length, so the $n+1$th is the shortest.  If the $n+1$th row is shorter than the 2nd, then some line that I have already created does not reach the line going from the leader through the $n+1$th row.  Therefore, all of the rows $b$ and on must be the same length.  
Since n is at least 2, there are at least three rows, so we can look at the last two rows to see that there must be $m^2$ lines between them, where m is the length of the last row (excluding the leader).  These must be the same lines as between the first two rows, and it is clear that there are $nm$ lines connecting the first two rows, so we see that $n=m$.  Therefore, it is clear that, if this is a finite projective plane, it is an $(n+1)n$ grid, along with a leader node.
