We use terminologies in graph theory. We shall prove by induction that in a complete graph of $3 \cdot k!$ vertices, if we colour every edge in one of $k$ colours, then there exists a monochromatic triangle.

For the base case $k=1$, since there are $3$ vertices but only one colour, the result holds trivially.

Assume the claim holds for some $k$. Consider a complete graph of $3 \cdot (k+1)!$ vertices. We colour every edge in one of $k+1$ colours. Fix any vertex $A$. By the pigeonhole principle, there are
$$\left\lfloor \frac{3 \cdot (k+1)! - 1}{k+1} \right\rfloor = 3 \cdot k!$$
edges with $A$ as an endpoint which are of the same colour, say red. If an edge connecting two of these vertices is red, then these two vertices together with $A$ form a red triangle. If all these edges are not red, then they are in one of $k$ colours. By the inductive hypothesis, there is a monochromatic triangle in this subgraph. This proves the inductive step, and hence the problem is solved.