What's the reason that exponentials are often used when requiring compact support? What's the reason that exponentials are often used when requiring compact support?
Such as
$$f_j=\frac{1}{j} e^{-\frac{1}{1-(j^2x)^2}}, |x| < 1/j^2$$
$0$ otherwise.
This will have a compact support.
However, I would like to understand what relevance does using $e^x$ have to compact support?

This is a kind of bump function I believe. Although I'm not very familiar with the use cases of bump functions.
https://en.wikipedia.org/wiki/Bump_function
 A: Not just compact support - $C^{\infty}$ with compact support. That means that the functions can't be analytic - because if they were, we could just continue them to be zero everywhere based on their power series somewhere outside that support. On the other hand, all of our formulas are analytic, so we need a boundary where we switch from some nonzero function with a usable formula to the zero function. At that boundary, the function and all of its derivatives must be zero in order for the two sides to agree.
So then, we need a function $f$ which is zero at some particular point - say, zero - and also with all of its derivatives zero at that point, while being definitely nonzero on one side - say, $x>0$.
First off, such an $f$ must have $\lim_{x\to 0^+}\frac{f(x)}{x^n} = 0$ for all $n$; it decays faster than any power of $x$. What do we know that's stronger than powers of $x$? Exponentials - or, at least, exponentials at $\pm\infty$. Near zero, that'll have to be $\exp\left(\pm\frac1x\right)$ instead. We want the one that decays for positive $x$, so that'll be $\exp\left(-\frac1x\right)$ for $x>0$.
This can be verified; $f(x)=\begin{cases}0&x\le 0\\ \exp\left(-\frac1x\right)&x>0\end{cases}$ is indeed $C^{\infty}$ on all of $\mathbb{R}$. It's not compactly supported because the domain of the second formula goes off to $\infty$, but that's easily fixed - just multiply by $f(1-x)$ to get
$$g(x)=f(x)f(1-x)=\begin{cases}0&x\le 0\\ \exp\left(-\frac1x-\frac1{1-x}\right)&0<x<1\\ 0&x\ge 1\end{cases}=\begin{cases}\exp\left(\frac{-1}{x-x^2}\right)&0<x<1\\ 0&\text{ otherwise}\end{cases}$$
You wrote down a more general form, but the only difference is a more general interval and a constant scale factor.
Basically, we use the exponentials because that's the simplest example we can write down.
And then - once we have one, we can build whatever we need. Take the convolution with a (not smooth) compactly supported function, and we get a $C^{\infty}$ compactly supported function. That convolution lets you do pretty much anything you need, to the point that we never actually need to find any other formulas for bump functions.
