Just some context: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times}\leq n$, means that any colouring of a complete graph, $k_n$, on $n$ vertices or more with $k$ colours must contain a monochromatic triangle.
Claim: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$
Proof:
Consider the minimum number of monochromatic edges connected to an arbitrary vertex, $V$. At least $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ edges will be monochromatic. $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ =$\frac{3k!}{k}$ when $k>1$. Therefore, vertex $V$ must be connected to $3(k-1)!$ monochromatic edges.
By trying to avoid a monochromatic triangle of color $k_1$, we will be forced into creating a monochromatic triangle of color {$k_2$, $k_3$, $\ldots$, or $k_i$}. If we connected any of the vertices $\{1,2,3, \ldots, 3(k-1)!\}$ with a blue edge, a blue monochromatic triangle would be formed, to avoid this, the remaining $3(k-1)!$ vertices must be colour with the remaining $k-1$ colours.
We can then consider a subgraph of the remaining vertices $\{1,2,3,\ldots,3(k-1)!\}$ and colour it with the remaining $k-1$ colours. We can focus on a new vertex, $V'$, within the subgraph.
Vertex $V'$ will be connected to at least $\displaystyle{\frac{3(k-1)!}{k-1}}$ monochromatic edges. Which equals $3(k-2)!$ edges.
We can continue this process until we consider the subgraph of remaining vertices $3(k-(k-1))!$ and colour it with the remaining $k-(k-1)$ colours. This means that this final subgraph has 3 vertices and has to be coloured with 1 colour. Therefore, a monochromatic triangle is unavoidable. Therefore, $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$.
I managed to get it down to $3k!$, I was wondering if anyone had better approximations?
Best Answer
You can simplify(?) your argument by using induction:
Let $f(k)=R(\underbrace{3,3,\ldots,3}_k)$.
In any $k$-colouring ($k\ge1$) of a $K_n$ graph ($n\ge2$), if we pick any vertex $v$, then by the pigeon-hole-principle, one of the $k$ colours ("blue", say) will occur for at least $m:=\left\lceil\frac{n-1}{k}\right\rceil$ times among the edges incident with $v$. If any edge between the other ends of these edges is blue, we have a blue triangle and are done. And if none of these edges is blue, we have found a subgraph that is a $(k-1)$-coloured $K_m$. Hence if $m\ge f(k-1)$ we are done again. In other words, we almost immediately find a monochromatic triangle if $\frac {n-1}k> f(k-1)-1$, or equivalently if $n>kf(k-1)-k+1$. We conclude that $$f(k)\le kf(k-1)-k+2.$$ Clearly, $f(1)=3$. From this, $f(2)\le 6$, $f(3)\le 17$, $f(4)\le 66$, etc. Thus we have found an upper bound $$ f(k)\le a_k$$ where $a_k$ is defined by the recursion $$\begin{align}a_1&=3,\\ a_{k}&=ka_{k-1}-k+2\qquad\text{for }k\ge2.\end{align}$$
Claim. We have $a_k-1=\sum_{i=0}^k\frac{k!}{i!}$.
Proof. This is clear for $k=1$. For $k\ge 2$ we find by induction $$a_k-1=k(a_{k-1}-1)+1=k\sum_{i=0}^{k-1}\frac{(k-1)!}{i!}+\frac{k!}{k!}=\sum_{i=0}^k\frac{k!}{i!}$$ $\square$
Corollary. $$R(\underbrace{3,3,\ldots,3}_k)\le \left\lceil k!e\right\rceil$$
Proof. This follows from $e=\sum_{i=0}^\infty\frac1{i!}$ and that the error $\sum_{i=k+1}^\infty\frac1{i!}$ is between $0$ and $1$ for $k\ge1$. $\square$
Remark. See also A073591.