The German Wikipedia is full of articles like this one saying one function falls slower than another. I wrecked google, but could not find the definition or sufficient conditions. Is it:
$ \lim_\limits{x \to \alpha }\frac{f(x)}{g(x)} = 0$, or
$\displaystyle{\frac{d}{dx}\frac{f(x)}{g(x)} < 0} $ (in appropriate region)
Or is any of these sufficient? I guess the first one, since it doesn't require differentiability.
Let's restrict this discussion to those stretched exponentials in the article you linked to, since they have the property that their limit is zero as $t \to +\infty$, which makes talking about the concept of "falling" relatively straightforward.
If we have two such functions, $f$ and $g$, then to say "$g$ falls more slowly than $f$" means that $g$ approaches $0$ more slowly than $f$ does. In other words, $f$ falls (decreases) more quickly than $g$ does. So it is the case that $\displaystyle\lim_{t\to +\infty} \frac{f(t)}{g(t)} = 0$, because the numerator approaches $0$ more quickly than the denominator does. There are two ways of stating this here: $$ \lim_{t \to+\infty} \frac{f(t)}{g(t)} = 0 \qquad \text{and} \qquad \lim_{t\to+\infty} \frac{g(t)}{f(t)} = +\infty$$
In general, $g(t) = e^{-t^\beta}$ falls more slowly than $f(t) = e^{-t}$ if we have $\beta < 1$. That's just from the link you provided. Consider: $$ \lim_{t\to +\infty} \frac{g(t)}{f(t)} = \lim_{t\to +\infty} \frac{e^{-t^\beta}}{e^{-t}} = \lim_{t\to+\infty} \frac{e^{-t^\beta} \cdot e^t}{e^{-t} \cdot e^t} = \lim_{t\to+\infty} e^{t-t^\beta}$$
Since $\beta < 1$ and $t \to +\infty$, then $t^\beta \to 0$. Thus $t - t^\beta \to +\infty$ as $t \to +\infty$, and so $\displaystyle \lim_{t\to+\infty} e^{t-t^\beta} = +\infty$. This paragraph isn't exactly rigorous, but I'm going for clarity here.
A similar argument shows the other limit mentioned above: $\displaystyle\lim_{t\to+\infty} \frac{f(t)}{g(t)} = 0$