Prove that $R_{k}(3,...,3) \leq (k+1)!$ for every integer $k \geq 2$ We've been given this problem in an old exam paper and I'm really struggling to see how it is done.
The first part asks us to define $R_{k}(s_1,s_2,...,s_k)$, and the second part asks us to prove that $R(3,3)=6$ which are both fairly straightforward.
My friend and I attempted to solve it by induction;
First, it holds on $k=2$ as $R(3,3)=6=(2+1)!$, so we proceed by assuming that it holds for;
$$R_{k-1}(3,...,3) \leq k!$$
But we're stuck from here and not sure what to do...
 A: We use notation $R_k(a)=R(\underbrace{a,a\dots,a}_k)$.
We prove $R_k(3)\leq (k+1)(R_{k-1}(3)-1)+2$.
We just have to prove any $k$ colored complete graph with at least that number of vertices has a monochromatic triangle.
Pick such a a graph $G$ and pick a vertex $g$, there is a color $c$ such that at least $R_{k-1}(3)+1$ of the edges coming from $v$ are of color $c$. Consider the set $C$ consisting of the  vertices of those edges other than $g$, if there is an edge of color $c$ between vertices in $C$ then a triangle of color $c$ is formed with $g$ and the ends of that edge. Otherwise the edges in $C$ only contain the other $k-1$ colors and since $|C|> R_{k-1}$a monochromatic triangle exists between vertices of  $C$.
We have thus proved that $R_k(3)\leq (k+1)(R_{k-1}(3)-1)+2$ for any $k\geq 3$. And since $R_{k-1}\geq 6$ we have $(k+1)(R_{k-1}(3)-1)+2= (k+1)R_{k-1}(3)-(R_{k-1}-2)\leq (k+1)R_{k-1}(3)$.
Therefore $R_k(3)\leq (k+1)R_{k-1}(3)$ for any $k\geq 3$.
So to prove $R_k(3)\leq (k+1)!$ we just do induction on $k$, equality occurs for $k=2$.
For the inductive step notice $R_k(3)\leq (k+1)!\implies (k+2)R_k(3)\leq (k+2)(k+1)!$. Finally complete this with:
$R_{k+1}(3)\leq(k+2)R_k(3)\leq (k+2)(k+1)!= (k+2)!$
