[gthompson512]

> - "My name is Hardy, G.H. Hardy.": A unique function satisfies exp(x+y) = exp(x)exp(y).

This has nothing to do with e and is satified by 2^x or any a^x, so this wouldn't work for introducing e in particular.

- "The Classic": There exists a unique function equal to its own derivative up to a constant.

Same for this, but if you fix the constant to be 1, then e^x is the only one that works.

I will give the series works too.

[qsort]

> This has nothing to do with e and is satified by 2^x or any a^x, so this wouldn't work for introducing e in particular.

You need to impose f'(0) = 1. (If you want to be really technical, also at least some regularity condition, I'll be honest, I don't remember what's the minimal one, let's say continuity)

> Same for this

I did say up to a constant.