How do I get this upper bound for Ramsey numbers: $R_k \le \left \lfloor k!e \right \rfloor + 1$?

Question

For every integer $k \ge 2$,

$$R_k \le \left \lfloor k!e \right \rfloor + 1$$

where $R_k$ denotes $R(\underbrace{{3, 3, \ldots, 3}}_{k})$.

Answer 1

The following text is quoted from the paper R. E. Greenwood, A. M. Gleason: Combinatorial relations and chromatic graphs, Canad. J. Math. 7(1955), 1-7, DOI: 10.4153/CJM-1955-001-4.

5. Upper and Lower Bounds for $n(3^r)$. Let $t_r=n(3^r)=n(3,3,\dots,3)$. The upper bounds used in Theorems 1, 4 and 5 can all be obtained by the use of

THEOREM 6. $$t_{r+1} \le (r+1)(t_r-1)+2.$$

This theorem is easily proved by induction; and then it is trivial to establish, also by induction, that $$t_{r+1} \le 3(r+1)!.$$ A somewhat sharper inequality may be obtained, however, without any added difficulties. It has already been established that $$t_r \le \lfloor (r!)e \rfloor +1, \qquad r=2,3,4,$$ where $\lfloor M \rfloor$ means the greatest integer contained in $M$. Such a bound holds for all integers $r\ge 2$, for if it did not there would be a least integer, say $s+1$, for which the relation failed to hold. By Theorem 6, $$t_{s+1} \le (s+1) \lfloor (s!)e \rfloor+2, \qquad s\ge2$$ But $\lfloor (s+1)!e \rfloor = (s+1) \lfloor (s!)e \rfloor +1$, and hence $$t_{s+1} \le \lfloor (s+1)!e \rfloor +1$$ and the stated upper bound follows.

A proof of the inequality stated as Theorem 6 can also be found here: Ramsey Number Inequality: $R(\underbrace{3,3,...,3,3}_{k+1}) \le (k+1)(R(\underbrace{3,3,...3}_k)-1)+2$