[Dove]

Great. Now I'm going to spend the rest of the morning trying to figure out whether you can really do it that way.

[Edit: Ah yes. The formalism goes like this: Consider the above series. By the above argument, it differentiates to itself. QED. ]

[tome]

Ah, but how do you know that e^x is the only thing that differentiates to itself? :)

[philh]

You don't need to. The uniqueness part of the definition is never used in that argument. (In fact, a*e^x also differentiates to itself, for any a; but that's a trivial case.)

Uniqueness almost follows from that argument. It's now easy to see that exp is the only analytic function satisfying exp' = exp and exp(0) = 1: if you have another one, by the same argument, it has the same Maclaurin expansion, hence is the same function.

However, I don't know how to prove uniqueness over all functions, not just analytic ones.