[proofplan] We induct on the number of colours $k$. The base case $k = 2$ is [Ramsey's Theorem](/theorems/2046). For the inductive step, we collapse a $k$-colouring into a 2-colouring by fusing colours $2, \ldots, k$ into a single new colour "not-1", then apply the two-colour Ramsey theorem to produce either a $K_{s_1}$ in colour 1 (which finishes the job) or a large monochromatic-in-"not-1" clique on which the original $k$-colouring uses only colours $2, \ldots, k$; the inductive hypothesis applied to this clique produces a $K_{s_i}$ in some colour $i \geq 2$. The critical size choice is $N = R\bigl(s_1, R_{k-1}(s_2, \ldots, s_k)\bigr)$. [/proofplan] [step:Induct on $k$ with base case the two-colour Ramsey theorem] We prove by induction on $k \geq 2$ the statement: for all $s_1, \ldots, s_k \geq 2$, there exists a positive integer $N$ such that every colouring $c : E(K_N) \to \{1, \ldots, k\}$ contains a monochromatic $K_{s_i}$ in colour $i$ for some $i \in \{1, \ldots, k\}$. The least such $N$ is $R_k(s_1, \ldots, s_k)$. **Base case $k = 2$.** For all $s_1, s_2 \geq 2$, $R_2(s_1, s_2) = R(s_1, s_2)$ exists by [Ramsey's Theorem](/theorems/2046), identifying colour $1 \mapsto \text{red}$ and colour $2 \mapsto \text{blue}$. [guided] The claim is that $R_k(s_1, \ldots, s_k)$ exists — i.e. there is some finite $N$ making every $k$-colouring of $K_N$ contain a monochromatic clique of the desired size in at least one of the colours. We prove existence; the least such $N$ is then well-defined by the well-ordering of $\mathbb{N}$. **Why induct on $k$?** The two-colour case is known (Ramsey's theorem), and the natural reduction to bring to bear is "merge colours". If we lump two colours together into a single colour, a $k$-colouring becomes a $(k-1)$-colouring — and we can use the inductive hypothesis on the smaller number of colours. **Base case.** When $k = 2$, there is nothing to prove beyond citing [Ramsey's Theorem](/theorems/2046): the two-colour Ramsey number $R(s_1, s_2)$ exists, and by renaming the colours $\{1, 2\} = \{\text{red}, \text{blue}\}$ we get $R_2(s_1, s_2) = R(s_1, s_2)$. [/guided] [/step] [step:Define the threshold $N = R(s_1, R_{k-1}(s_2, \ldots, s_k))$ using the inductive hypothesis] Suppose $k \geq 3$ and that the theorem holds for $k - 1$ colours. Define \begin{align*} M &= R_{k-1}(s_2, \ldots, s_k), \\ N &= R(s_1, M), \end{align*} both of which exist: $M$ exists by the inductive hypothesis applied to the $k - 1$ colours $\{s_2, \ldots, s_k\}$ (each $s_i \geq 2$), and $N$ exists by [Ramsey's Theorem](/theorems/2046) since $s_1 \geq 2$ and $M \geq 2$ (because $M \geq R(s_2, s_3) \geq \max(s_2, s_3) \geq 2$). We claim that every colouring $c : E(K_N) \to \{1, \ldots, k\}$ contains a $K_{s_i}$ in colour $i$ for some $i$, which establishes $R_k(s_1, \ldots, s_k) \leq N$ and hence its existence. [guided] We need to pick $N$ large enough that any $k$-colouring of $K_N$ forces a monochromatic $K_{s_i}$ in some colour $i$. The idea: 1. Collapse colours $2, \ldots, k$ into one meta-colour "not-1". This creates a 2-colouring of $K_N$. 2. If the original 2-colour Ramsey theorem on this 2-colouring produces a "colour-1" clique of size $s_1$, we are done: the "colour-1" edges are literally the original colour-1 edges, so this is a $K_{s_1}$ in colour 1. 3. If instead it produces a "not-1" clique of some target size $M$, then within this clique every edge is of colour in $\{2, \ldots, k\}$. We now have a $(k - 1)$-colouring of $K_M$, and we want $M$ chosen so that this forces a $K_{s_i}$ in colour $i$ for some $i \in \{2, \ldots, k\}$. So we choose $M = R_{k-1}(s_2, \ldots, s_k)$ (exists by induction) and $N = R(s_1, M)$ (exists by the two-colour theorem). We must check the hypotheses of Ramsey's theorem: both arguments must be at least $2$. $s_1 \geq 2$ is given. $M = R_{k-1}(s_2, \ldots, s_k) \geq 2$ holds because $M$ is at least the size of a $K_{s_2}$, and $s_2 \geq 2$, so $M \geq s_2 \geq 2$. Concretely $M \geq R(s_2, s_3) \geq \max(s_2, s_3) \geq 2$. [/guided] [/step] [step:Collapse to two colours and apply the two-colour Ramsey theorem] Given a colouring $c : E(K_N) \to \{1, \ldots, k\}$, define the "collapsed" 2-colouring \begin{align*} \tilde{c} : E(K_N) &\to \{1, \ast\} \\ e &\mapsto \begin{cases} 1 & \text{if } c(e) = 1, \\ \ast & \text{if } c(e) \in \{2, \ldots, k\}, \end{cases} \end{align*} where $\ast$ is a formal symbol representing "not colour 1". By definition of $N = R(s_1, M)$ via [Ramsey's Theorem](/theorems/2046), the 2-colouring $\tilde{c}$ contains either - a $K_{s_1}$ with all edges coloured $1$ under $\tilde{c}$, or - a $K_M$ with all edges coloured $\ast$ under $\tilde{c}$. **Subcase 1.** A $K_{s_1}$ on $W \subseteq V(K_N)$ with all edges $\tilde{c}$-coloured $1$. By the definition of $\tilde{c}$, the $\tilde{c}$-preimage of $1$ is exactly the $c$-preimage of $1$; hence every edge of $K_N[W]$ satisfies $c(e) = 1$. We have exhibited a $K_{s_1}$ of colour $1$, and the theorem holds. **Subcase 2.** A $K_M$ on $U \subseteq V(K_N)$ with all edges $\tilde{c}$-coloured $\ast$. By the definition of $\tilde{c}$, the $\tilde{c}$-preimage of $\ast$ is exactly the $c$-preimage of $\{2, \ldots, k\}$. Hence the restriction \begin{align*} c|_{E(K_N[U])} : E(K_N[U]) \to \{2, \ldots, k\} \end{align*} takes values in the $k - 1$ colours $\{2, \ldots, k\}$. Relabelling $\{2, \ldots, k\}$ as the $k - 1$ colour classes of a $(k-1)$-colouring of $K_M$, the inductive hypothesis applied to $R_{k-1}(s_2, \ldots, s_k) = M = |U|$ produces a monochromatic $K_{s_i}$ of colour $i$ for some $i \in \{2, \ldots, k\}$. This is also a monochromatic $K_{s_i}$ of colour $i$ in the original colouring $c$ on $K_N$, finishing the inductive step. This completes the induction and proves that $R_k(s_1, \ldots, s_k)$ exists for all $k \geq 2$ and all $s_1, \ldots, s_k \geq 2$. [guided] **Defining $\tilde{c}$.** The collapsed 2-colouring $\tilde{c}$ identifies all of colours $2, \ldots, k$ as a single meta-colour $\ast$, leaving colour $1$ untouched: \begin{align*} \tilde{c}(e) &= \begin{cases} 1 & c(e) = 1 \\ \ast & c(e) \in \{2, \ldots, k\} \end{cases}. \end{align*} This is a well-defined map to a 2-element set $\{1, \ast\}$, and renaming $1 \mapsto \text{red}$, $\ast \mapsto \text{blue}$ shows it is a 2-edge-colouring. **Applying the two-colour Ramsey theorem.** By choice $N = R(s_1, M)$, the two-colour Ramsey theorem guarantees either a red $K_{s_1}$ or a blue $K_M$ under $\tilde{c}$. Here we verify its hypotheses: the theorem requires both parameters $\geq 2$; we have $s_1 \geq 2$ by assumption and $M = R_{k-1}(s_2, \ldots, s_k) \geq 2$ because $R_{k-1}(s_2, \ldots, s_k)$ is always at least $s_2 \geq 2$ (any monochromatic-in-$2$ $K_{s_2}$ requires at least $s_2$ vertices). **Unpacking each subcase.** *Subcase 1 — red $K_{s_1}$ in $\tilde{c}$.* The $\tilde{c}$-colour $1$ is defined precisely as the $c$-colour $1$: $\tilde{c}(e) = 1 \iff c(e) = 1$. So a $\tilde{c}$-monochromatic-$1$ clique is literally a $c$-monochromatic-$1$ clique. This gives a $K_{s_1}$ of $c$-colour $1$ — a finished instance of the theorem. *Subcase 2 — blue $K_M$ in $\tilde{c}$ on vertex set $U$.* Every edge of $K_N[U]$ is $\tilde{c}$-coloured $\ast$, hence $c$-coloured in $\{2, \ldots, k\}$. So $c$ restricted to $K_N[U] \cong K_M$ is a $(k-1)$-colouring using exactly the colours $\{2, \ldots, k\}$. Since $|U| = M = R_{k-1}(s_2, \ldots, s_k)$, the inductive hypothesis for $k - 1$ colours (with parameters $s_2, \ldots, s_k$) guarantees a monochromatic $K_{s_i}$ in some colour $i \in \{2, \ldots, k\}$. This clique, viewed inside $K_N$, is still monochromatic in the original colour $i$, and $i \geq 2$. **Why the induction closes.** In each subcase we have produced a monochromatic $K_{s_i}$ in some colour $i \in \{1, 2, \ldots, k\}$. Hence $R_k(s_1, \ldots, s_k) \leq N$, so $R_k(s_1, \ldots, s_k)$ exists. [/guided] [/step]