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

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.