n^2 is polynomial, not exponential. If it was truly exponential (which a sort should never be unless you've decided to solve an arbitrary 3-SAT problem while sorting a list), then it would've blown up in your face much sooner.
There's certainly a most awesome sort algorithm which is exponential...
https://www.dangermouse.net/esoteric/bogobogosort.html
They are not even sure what the complexity is but it's like O(n!^(n-k)) or O(n*(n!)^n).
"Exponential" is indeed misused.
Half the time I hear it, it is lower, like quadratic or cubic. The other half the time is higher, like combinatorial.
But even without any scaling factors, don't underestimate the size of the numbers involved. If you had two threads, one simply counting up to some n! upper bound, and the other up to 10^n - how long do you think the minimum problem size would take for the "slow" combinatorial to actually be slower than the exponentiation? Hint: at 4GHz and 1 op/cycle... Let's just say I'm not holding my breath that our species will still be around then. And even a tiny scaling factor to the mix...
I mean, if you're trying to estimate computational complexity at least, this nuance seems pointless.
But quadratic vs. exponential really matters, with plausible parameters.
I'm curious if the rise of quantum computers will make the differences between exponential and combinatorial meaningful in a practical sense.
This is why you sometimes see complexities that are e.g. O(1.3894732894^n) in wikipedia articles on the best known cases for various algorithms.
Typo: O(a^n) is a stronger guarantee than O(b^n) if a < b. It means nothing if a > b.
I actually thought the most likely intended meaning was that an algorithm in O(a^n) must take asymptotically longer than one in O(b^n) as n goes to infinity (if a > b), which isn't true. (For example, when a > b, then every algorithm in O(b^n) is also in O(a^n), but obviously no algorithm can asymptotically require more time than itself.)