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CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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Dobinski's formula expresses the Bell numbers as Bn=e10n/!

Statement

For every nN,

Bn=e10n!.

Facts & Assumptions

Given: A natural number n.

[F1]

Ordinary powers expand as n=k=0nS(n,k)k. (Ordinary powers expand in the falling-factorial basis by the second-kind Stirling numbers).

[F2]

The falling factorial satisfies k=0 for <k and k/!=1/(k)! for k (The factorial n! and the falling factorial nk, defined by recursion in N).

[F4]

The Bell number is Bn=k=0nS(n,k) (The Stirling numbers of the second kind and the Bell numbers).

Proof

technique · direct
1.1

For a fixed kn, [F2] and the change of index m=k give. [F2, F3, algebra] 0k!=k1(k)!=m01m!=e. In particular each of these nonnegative series converges.

F2F3algebra
2.1

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 0n!=k=0nS(n,k)0k!=ek=0nS(n,k)=eBn.

F1F4step 1.1algebra
3.1

Multiplying step 2.1 by e1 and using [F3] gives. [F3, step 2.1, algebra] Bn=e10n!. This also covers n=0, where 00=1 is the natural-power base convention already used in [F1].

F3step 2.1algebra

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