TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01

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Cayley's formula: $τ (K_{n}) = n^{n - 2}$ for $n \geq 2$ , with $τ (K_{1}) = 1$ and $τ (K_{0}) = 0$

Statement

For $n \geq 2$ , the complete graph $K_{n}$ has

$τ (K_{n}) = n^{n - 2}$

spanning trees. Moreover, $τ (K_{1}) = 1$ and $τ (K_{0}) = 0$ .

Facts & Assumptions

Given: A natural number $n$ .

[L1]

Trees on the fixed label set $n$ are in bijection with words of length $n - 2$ on $n$ when $n \geq 2$ (Prüfer encoding and decoding are inverse bijections between labelled trees on $n \geq 2$ vertices and words of length $n - 2$ on their labels).

[L2]

The set of functions from an $m$ -element set to an $n$ -element set has cardinality $n^{m}$ (The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $∣ A^{B} ∣ = ∣ A ∣^{∣ B ∣}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ ).

[F1]

Every labelled tree on $n$ is a spanning subgraph of $K_{n}$ (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices, The spanning-tree number $τ (G)$ ).

Proof

technique · direct

1.1

For $n \geq 2$ , spanning trees of $K_{n}$ are exactly the trees on its fixed label set.

2.1

By the Prüfer bijection, they correspond to functions from an $(n - 2)$ -element position set to $n$ , of which there are $n^{n - 2}$ .

step 1.1L1L2

3.1

The sole spanning tree of $K_{1}$ is $K_{1}$ itself, so $τ (K_{1}) = 1$ . The null graph $K_{0}$ is not a tree and has no spanning tree, so $τ (K_{0}) = 0$ .

F1∎

Depends on

Used by

Cayley's formula follows again from the matrix-tree theorem Corollary
τ(K₄)=16 by direct Prüfer enumeration and Cayley's formula Example
Cayley's formula counts unlabelled trees False statement

Dependency tree · two levels

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Sources

ISI Bangalore discrete mathematics notes, Trees and Cayley’s theorem (standard reference, not scraped)

Cayley's formula: τ(Kn)=nn−2τ(Kn​)=nn−2 for n≥2n≥2, with τ(K1)=1τ(K1​)=1 and τ(K0)=0τ(K0​)=0