CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Dobinski's formula expresses the Bell numbers as $B_{n} = e^{- 1} \sum_{ℓ \geq 0} ℓ^{n} / ℓ!$

Statement

For every $n \in N$ ,

$B_{n} = e^{- 1} \sum_{ℓ \geq 0} \frac{ℓ^{n}}{ℓ!} .$

Facts & Assumptions

Given: A natural number $n$ .

[F1]

Ordinary powers expand as $ℓ^{n} = \sum_{k = 0}^{n} S (n, k) ℓ^{\underline{k}} .$ (Ordinary powers expand in the falling-factorial basis by the second-kind Stirling numbers).

[F2]

The falling factorial satisfies $ℓ^{\underline{k}} = 0$ for $ℓ < k$ and $ℓ^{\underline{k}} / ℓ! = 1 / (ℓ - k)!$ for $ℓ \geq k$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[F3]

The real exponential is defined by $e = \sum_{m \geq 0} 1 / m!$ , and $e^{- 1} e = 1$ (The real exponential function and the number $e$ by a power series, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

[F4]

The Bell number is $B_{n} = \sum_{k = 0}^{n} S (n, k)$ (The Stirling numbers of the second kind and the Bell numbers).

Proof

technique · direct

1.1F2F3algebra

For a fixed $k \leq n$ , [F2] and the change of index $m = ℓ - k$ give. [F2, F3, algebra] $\sum_{ℓ \geq 0} \frac{ℓ^{\underline{k}}}{ℓ!} = \sum_{ℓ \geq k} \frac{1}{(ℓ - k)!} = \sum_{m \geq 0} \frac{1}{m!} = e .$ In particular each of these nonnegative series converges.

2.1F1F4step 1.1algebra

Substitute [F1] into the series in the Statement. Since the sum over $k$ . [F1, F4, step 1.1, algebra] is finite, it may be interchanged with the convergent nonnegative series, and step 1.1 yields $\sum_{ℓ \geq 0} \frac{ℓ^{n}}{ℓ!} = \sum_{k = 0}^{n} S (n, k) \sum_{ℓ \geq 0} \frac{ℓ^{\underline{k}}}{ℓ!} = e \sum_{k = 0}^{n} S (n, k) = e B_{n} .$

3.1F3step 2.1algebra∎

Multiplying step 2.1 by $e^{- 1}$ and using [F3] gives. [F3, step 2.1, algebra] $B_{n} = e^{- 1} \sum_{ℓ \geq 0} \frac{ℓ^{n}}{ℓ!} .$ This also covers $n = 0$ , where $0^{0} = 1$ is the natural-power base convention already used in [F1].

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Sources

Wolfram MathWorld, Dobiński's Formula (standard reference, not scraped)
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics — Symbolic Combinatorics (standard reference, not scraped)

Dobinski's formula expresses the Bell numbers as Bn=e−1∑ℓ≥0ℓn/ℓ!Bn​=e−1∑ℓ≥0​ℓn/ℓ!