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by tome 5890 days ago
Ah, but how do you know that e^x is the only thing that differentiates to itself? :)
1 comments

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.