TITLE: Is $e^x=\exp(x)$ and why?
QUESTION [10 upvotes]: In the comments to this question a discussion came up wether we have $e^x=\exp(x)$ by definition and what the "correct" definition of $\exp(x)$ is. Building on that, I want to line out the problem with this question and give one way to prove $e^x=\exp(x)$ for $x\in\mathbb R$. 
(Be warned: this is a long post)

REPLY [8 votes]: Usually one defines $\exp x=\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}$.
It is easy to show it converges for all $x$ (even for complex values) and it satisfies the differential equation with initial condition:
$$y'=y,\quad y(0)=1.$$
Now this differential equation implies the functional relation:
$$\exp(x+y)=\exp x\cdot\exp y$$
which in turn implies that for all $n\in\mathbf N$, $\;\exp(nx)=(\exp x)^n$, whence for any $\dfrac pq$, $\;\exp\Bigl(\dfrac pqx\Bigr)=\bigl(\exp x\bigr)^{\tfrac pq}$.
Now set $\mathrm e=\exp 1=\displaystyle\sum_{n=0}^{\infty}\frac1{n!}$. What precedes proves that for any rational $r$, we have:
$$\exp r = \mathrm e^r,$$
and we can define $\mathrm e^x$, for $x\in\mathbf C\smallsetminus \mathbf Q$, as:
$$\mathrm e^x\overset{\text{def}}{=} \exp x\quad\text {if}\enspace x\in\mathbf C\smallsetminus \mathbf Q.$$