I have been studying smooth infinitesimal analysis which depends entirely on the set $\Delta$ of numbers whose square (and all higher powers) are equal to $0$. Now, I know smooth infinitesimal analysis is a branch of constructive mathematics, and I was just wondering if the properties of exponents like,
$a^x \cdot a^y = a^{x+y}, (a^x)^y = a^{xy}, \text{ and } a^{1/n} = \sqrt[n]{a}$
would hold if $a \in \Delta$. Also, is this a valid proof of $\epsilon = 0$ where $\epsilon \in \Delta$ and $\epsilon \geq 0$:
Assume $\epsilon \geq 0$ is an element of $\Delta$. Clearly, $\epsilon^1=\epsilon^{2/2} = \sqrt{\epsilon^2}$. But, $\epsilon^2 = 0$ as $\epsilon \in \Delta$ and $x,y \in \mathbb{R}$, so we have $\epsilon = 0$.
I do not have a great amount of experience in constructive mathematics and have only learned smooth infinitesimal analysis because of its applications (like deriving the formula for the surface area of a sphere) up to this point, so any guidance on how to properly apply constructive mathematics would be useful. I am also new to the site, so I will gladly accept advice on how to improve my question.
Yes. In fact, it's quite common to take as an axiom:
(See, for instance, Axiom VI in O'Connor's notes)
Now, since $(x,y) \mapsto a^x \cdot a^y$ and $(x,y) \mapsto a^{x+y}$ are equal as smooth functions $\mathbb{R}^2 \to \mathbb{R}$, we get your corresponding identity as an axiom. The identity $(a^x)^y = a^{xy}$ is true for the same reason.
However, we have to be careful with $\sqrt{a}$, since this function isn't smooth! Indeed, even its first derivative isn't defined at $x=0$. We can recover, though, as long as we're working with numbers instead of functions. For instance, there is a number $c = 1.414\ldots$ so that $c^2 = 2$. Even though we can't write down a function $x \mapsto \sqrt{x}$, we can write down the equation $c^2 = 2$.
Indeed, if we view constants as smooth functions $c, 2 : \mathbb{R}^0 \to \mathbb{R}$, then the same axiom as before tells us that we have smooth functions $c, 2 : R^0 \to R$ (that is, members of $R$) synthetically too. Moreover, this axiom tells us that the externally true equation $c^2 = 2$ is true is true of these synthetic smooth functions as well! In this way, when working with numbers, we can cheat and write "$c = \sqrt{2}$", but we have to be careful to remember that this is really an abbreviation for the smooth statement $c^2 = 2$, rather than anything to do with any kind of function $x \mapsto \sqrt{x}$ (which, again, isn't smooth! So it doesn't exist synthetically).
With this caveat in mind, though, our usual statements $(a^{1/n})^n = a$ are now true for any fixed number $a$ basically by definition! Since we define $a^{1/n}$ to be shorthand for "the number $c$ so that $c^n = a$" just like we did with $\sqrt{2}$ earlier.
This also shows why your purported proof that $\epsilon = 0$ fails. Since $\epsilon$ isn't an (external) real number, the symbol "$\sqrt{\epsilon}$" doesn't actually mean anything inside our theory! After all, it can't possibly be an abbreviation for "some number $c$ so that $c^2 = \epsilon$" since there's no external number $\epsilon$ to apply our axiom to!
I hope this helps ^_^