*Dobinski’s formula* entails the following infinite series expression for the *n*th Bell number :

In this post we’ll work through a proof of Dobinski’s formula. We’ll need four formulas:

- The Maclaurin series for : .
- The formula for converting normal powers to falling factorial powers: where is a Stirling number of the second kind
*.*
- The formula relating Bell numbers and Stirling numbers of the second kind: . (This follows directly from the combinatorial definitions of the Bell numbers and the Stirling numbers of the second kind.)
- The representation of falling factorial powers as factorials: , valid when and .

Here we go… Starting with the infinite series expression convert to falling factorial powers to get

Thus .

(I made a similar argument in this Math.SE question.)

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very nice. thanks

I think #4 condition is not needed since if k<j, then k(k-1)(k-2)..(k-j+1)=0 since one of the factors (one of the round brackets) will be 0. Thanks for the proof!