I have noticed the following:
$$\left( (-1)^{1/2} \right)^2 = i^2= -1$$ But $$\left( (-1)^{2} \right)^{1/2} = \sqrt{1} = 1$$ I understand with complex numbers there is an issue of taking the principal root, yet this problem is still present. Under what circumstances will exponents commute?
On
Complex exponentiation is defined as $a^b=e^{b\log a}.$ To analyze the commutativity in this case, $i^2=e^{2\log i}$ and $\log i =\ln(1)+i(2\pi m+\pi/2)$ and $m\in\Bbb{Z}$ depends on the branch of logarithm that we choose. Therefore $$i^2=e^{2i(2\pi m+\pi/2)}=e^{i\pi}=-1$$ regard less of the branch. On the other hand, $\sqrt{1}=e^{1/2\log 1},$ where $\log 1=\ln(1)+2\pi n i$ for some $n\in\Bbb{Z}.$ Thus $$\sqrt{1}=e^{n\pi i}$$ and this depends on the branch that we choose. According to the parity of $n,$ here we have $$\sqrt{1}=\pm1.$$
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You don't really need to think about complex numbers here. Just look at $$ \left((-1)^2\right)^{1/2} = 1 $$ and $$ (-1)^1 = -1. $$ The issue here is the fact that the equation $x^2=1$ has two solutions, $x=+1$ and $x=-1$. When you write $1^{1/2}$, you're taking the positive square root of $1$, which is $1$.
To elaborate
Let's rewrite your problem as follows: $$ \textbf{Problem 1: }\text{Find $x$ such that } x = \left((-1)^2\right)^{1/2}. $$ Look at the related problem: $$ \textbf{Problem 2: }\text{Find $x$ such that } x^2 = (-1)^2. $$ Clearly, Problem 2 has two solutions, $+1$ and $-1$. Problem 1 is essentially asking for the positive solution to Problem 2, hence the inconsistency.
In short: As John Lou said in his comment, if the base is negative, you can't necessarily commute exponentiation.
Integer powers commute. If $a,b\in \Bbb Z$, then $(z^a)^b=z^{ab}=(z^b)^a$, for any complex $z$. Further, if $x$ is a positive real number, any real powers commute, so $(x^a)^b=x^{ab}=(x^b)^a$ for $a,b\in\Bbb R$.
Outside of those special cases, it’s best to be careful.