Is it true that $\forall s \geq 2, \forall k \geq 2,\; R_k(s) \leq (R_{k-1}(s))^s$? Here we denote by $R_k(s)$ the least integer $n$ such that $k$-colouring $K_n$ yields a monochromatic $K_s$
In order to prove another result ($R_k(s) \leq 4^{s^{k-1}}$) I would like to prove the result in the title: $ R_k(s) \leq (R_{k-1}(s))^s$
However, I am not sure how I may begin to do this. My first thought is that: 
$\forall k \geq 2, \; R_k(s) \geq R_2(s) $ so if we let $n = R_k(s)$ and consider 2-colouring $K_{n^s}$, then we know we get a monochromatic $K_s$. If we call the colours blue and yellow, with yellow representing $k-1$ colours other than blue, then we either get a blue $K_s$ or a yellow $K_s$.
If its a blue $K_s$ then we are done, so we consider what happens if we have a yellow $K_s$.
It is at this point that I realise that a $K_s$ is too small for my goal and that if I can instead claim that I have a monochromatic $K_n$ then I would be done. However, I do not know how I can justify that this is the case. 
This would necessitate that $R_2(n) \leq n^s$. I have that $R_2(s) \leq 4^s$ and so $R_2(n) \leq 4^n$ but I can't see why it would be true, if at all, that $4^n \leq n^s$, i.e. $4^{R_k(s)} \leq {R_k(s)}^s$
At this point I am completely unsure of how to continue and would appreciate a hint in the right direction, or at least verification of whether or not what I'm thinking can actually yield the desired result in the title. If there is another way to prove the inequality mentioned at the start then that would also be appreciated, thank you in advance. 
 A: Consider a coloring $\chi$ of $K_N$ with $k$ colors and no $K_{s_i}$ in color $i$. Pick your favorite vertex $v\in V(K_N)$ and partition $V(K_N)\setminus\{v\}$ into $V_1,\dots,V_k$ where $V_i=\{u\in V(K_N):\chi(v,u)=i\}$. By assumption, we must have $|V_i|\leq R(s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k)-1$ (the off-diagonal multicolor Ramsey number). In particular, we have $$R(s_1,\dots,s_k)\leq\sum_{i=1}^k\bigl(R(s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k)-1\bigr)+1\leq \sum_{i=1}^kR(s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k).$$
Now, I want to claim that if we have non-negative integers $s_1,\dots,s_k$ with $s_1+\dots+s_k=n$ then $${n\choose s_1,\dots,s_k}=\sum_{i=1}^k{n-1\choose s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k}.$$ Indeed, this is a quick double-counting exercise. 
We now prove that $R(s_1,\dots,s_k)\leq{\sum_i s_i\choose s_1,\dots,s_k}$. For a starting place, notice that, trivially, $R(1,1,\dots,1)=1={k\choose 1,1,\dots,1}$. By the previous remarks and induction $$R(s_1,\dots,s_k)\leq \sum_{i=1}^kR(s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k)\leq \sum_{i=1}^k{\bigl(\sum_i s_i\bigr)-1\choose s_1,\dots,s_{i-1},s_i-1,s_{i+1},\dots,s_k}={\sum_i s_i\choose s_1,\dots,s_k}.$$
In particular, we've shown that $R_k(s)\leq{ks\choose s,s,\dots,s}$. For the final (poor) bound, we find $${ks\choose s,s,\dots,s}\leq\sum_{r_1+\dots+r_k=ks}{ks\choose r_1,\dots,r_k}=k^{ks}=4^{sk\log_4 k},$$ which is better than you were hoping for.
