Im trying to prove that for any number n the complete graph with $p(n)$ vertices whose edges have been colored with n colors in some way has a monochromatic triangle (a triplet of nodes that are connected with edges of the same color) regardless of the way the coloring has been made.
$$p(n) = \lfloor (e*n!) \rfloor+1$$
I have proven the statement using pigeonhole principle for $n=1,2,3$.
For $n=2$, $p(n) = 6$. We pick a node and it has $5$ outgoing edges. From pigeonhole principle at least $3$ are colored the same, say red. Now let $A,B,C$ be the receiving vertices of these edges. If there is at least one red edge in $ABC$ were done. If not were still done.
For $n=3$, $p(n) = 17$. Same logic. Pick a node, $16$ edges spawn from it. At least $6$ (notice that this is $p(2)$) have the same color and the receiving nodes of these 6 edges form a complete graph with $p(2)$ nodes. The problem is recursively transformed into something we have already solved. For the general case i try to apply the same logic: pick a node. From there $p(n)-1$ edges are left.
$\lceil (p(n)-1)/n\rceil$ of those edges are the same color, say red. If in that smaller graph a red edge exists we're done. If not, we have $n-1$ colors to paint the edges of a complete graph with $\lceil(p(n)-1)/n \rceil$ nodes. This can be done if $ \lceil(p(n)-1)/n \rceil = p(n-1)$. From here i want to transform this recursion into the formula i gave in the problem description. Any ideas?