[proofplan] To show $R(s) > (s - 1)^2$, it suffices to exhibit a 2-colouring of the edges of $K_{(s-1)^2}$ with no monochromatic $K_s$. We partition the vertex set into $s - 1$ groups of $s - 1$ vertices, colour *intra-group* edges blue and *inter-group* edges red, and verify each constraint separately. The absence of a red $K_s$ uses pigeonhole: any $s$ vertices among $s - 1$ groups must have two in the same group, whose edge is blue. The absence of a blue $K_s$ uses the group-size bound: a blue clique lies inside one group, which holds only $s - 1 < s$ vertices. [/proofplan] [step:Exhibit the construction: partition vertices into $s - 1$ groups and colour] Let $n := (s - 1)^2$ and let $V := \{1, 2, \ldots, n\}$ be the vertex set of $K_n$. Partition $V$ into $s - 1$ pairwise disjoint blocks $V_1, \ldots, V_{s-1}$, each of size $s - 1$: \begin{align*} V_k := \{(k-1)(s-1) + 1,\ (k-1)(s-1) + 2,\ \ldots,\ k(s-1)\}, \qquad 1 \leq k \leq s - 1. \end{align*} Then $|V_k| = s - 1$ for each $k$ and $V = V_1 \sqcup \cdots \sqcup V_{s-1}$. Define a 2-colouring \begin{align*} c: V^{(2)} &\to \{\text{red}, \text{blue}\} \\ \{u, v\} &\mapsto \begin{cases} \text{blue} & \text{if } u, v \in V_k \text{ for some } k, \\ \text{red} & \text{otherwise (i.e.\ $u$ and $v$ lie in different blocks).} \end{cases} \end{align*} The two cases are mutually exclusive and exhaustive, so $c$ is well defined. We verify that $c$ has no monochromatic $K_s$. [/step] [step:Rule out a monochromatic red $K_s$ by pigeonhole across blocks] Suppose, for contradiction, $S \subseteq V$ with $|S| = s$ is a red clique under $c$, i.e., $c(\{u, v\}) = \text{red}$ for every $\{u, v\} \in S^{(2)}$. Consider the function $\pi : S \to \{1, \ldots, s - 1\}$ sending $v \in S$ to the unique index $k$ with $v \in V_k$. Since $|S| = s$ and $|\{1, \ldots, s - 1\}| = s - 1$, by the pigeonhole principle there exist distinct $u, v \in S$ with $\pi(u) = \pi(v) = k$. Thus $u, v \in V_k$, so by definition of $c$ the edge $\{u, v\}$ is blue, contradicting that $S$ is a red clique. Hence no red $K_s$ exists. [guided] The red colour class encodes "inter-block" edges. Any $s$ vertices chosen from the $s - 1$ blocks must, by the pigeonhole principle applied to the block-assignment function $\pi : S \to \{1, \ldots, s - 1\}$, include at least two vertices from a common block. But our colouring rule makes that pair blue — not red. So no $s$-clique can be entirely red. Concretely, with $|S| = s > s - 1 = $ number of blocks, the map $\pi$ cannot be injective. Let $u \neq v$ with $\pi(u) = \pi(v) = k$; then $u, v \in V_k$ and $c(\{u, v\}) = \text{blue}$. This single blue edge prevents $S$ from being red-monochromatic. The size $s - 1$ for the number of blocks is precisely what makes pigeonhole fire at $|S| = s$: with $s - 1$ blocks and $s$ vertices, a collision is forced. This is why we chose $s - 1$ blocks (not $s$, not $s - 2$). [/guided] [/step] [step:Rule out a monochromatic blue $K_s$ by the block-size constraint] Suppose, for contradiction, $S \subseteq V$ with $|S| = s$ is a blue clique under $c$. Pick any $u \in S$ and let $k$ be the unique index with $u \in V_k$. We claim $S \subseteq V_k$. Indeed, for any other $v \in S$, the edge $\{u, v\}$ is blue by hypothesis. By definition of $c$, a blue edge requires $u$ and $v$ to lie in the same block, so $v \in V_k$. Thus $S \subseteq V_k$. But $|S| = s$ and $|V_k| = s - 1 < s$, which contradicts $S \subseteq V_k$. Hence no blue $K_s$ exists. [/step] [step:Conclude $R(s) > (s - 1)^2$] We have exhibited a 2-colouring of the edges of $K_{(s-1)^2}$ containing no monochromatic $K_s$. By the definition of the Ramsey number $R(s) = R(s, s)$ as the smallest $n$ such that every 2-colouring of $K_n$ contains a monochromatic $K_s$, this shows $R(s) > (s - 1)^2$. [/step]