$\exp(q)=e^q, q\in\mathbb{Q}$

Question

Let $\exp(x):=\sum_{n=0}^{\infty} \frac{x^n}{n!}$ and $e:=\exp(1)$. I've managed to prove that $\exp(x)=e^x$ when $x$ is natural or integer and that $\exp(nx)=(\exp(x))^n$ and now I want to show (using only basic properties of $\exp$ and without using logarithms) that $\exp(q)=e^q\ , q\in\mathbb{Q}$. I've tried but without success so I'd appreciate any hint.

Answer 1

Usually I recommend to do it this way:

Hint 1: Try to prove $\exp(a+b)=\exp (a)\cdot\exp (b)$ first. The rest is then straight forward.

---

Or given what you already have, try this:

Hint 2: If $q=1/n$, then you have $e=\exp(1)=\exp(1/n\cdot n)=\exp(1/n)^n$, and this implies $$\exp(1/n)=\sqrt[n]{e}=e^{1/n}.$$

Can you find the solution for general $q\in\Bbb Q$?