First of all, kudos for introducing these ideas to your 10 year-old! That's an excellent way to get them interested in mathematics at an early age.

As to your question: The short answer is no. Any algebraic operation that you can do of the sort you're describing will yield a complex number. This is due to the fact that they are *algebraically closed*. What this means is the following.

One way that you can show that you can find a larger domain than the real numbers is by looking at the polynomial
$$
f(x) = x^2 + 1
$$
You can easily see that there are no solutions to the equation $f(x) = 0$ in the real numbers (just as the equation $x + 1 = 0$ has no solutions in the positive integers). So we must escape to a larger domain, the complex numbers, in order to find solutions to this equation.

A domain (to use your term) being algebraically closed means that *every* polynomial with coefficients in that domain has solutions in that domain. The complex numbers are algebraically closed, so no matter what polynomial-type expression that you write down, it will have as solution a complex number.

Now, this isn't to say that there aren't larger domains than $\mathbb{C}$! One example is the *Quaternions*. Where the complex numbers can be visualized as a plane (i.e. $a + bi \leftrightarrow (a, b)$), the quaternions can be visualized as a four-dimensional space. These are given by things that look like
$$
a + bi + cj + dk
$$
where $i, j, k$ all satisfy $i^2 = j^2 = k^2 = -1$, and moreover $ij = -ji = k$. The interesting fact about the quaternions is that they are *non-commutative*. That is, the order in which we multiply matters!

There are also Octonions, which are even weirder, and are an 8-dimensional analogue.

Anyhow, the answer is in the end that it sort of depends. In most senses, the complex numbers are as far as you can go in a relatively natural way. But we can still look at bigger domains if we want, but we have to find other ways to build them.

You can put (partial) orders on the complex numbers. One choice is to compare the real parts and ignore the complex ones. Another is to use the lexicographic order, comparing the real parts and then comparing the imaginary ones if the real parts are equal. Another is to use the modulus. There are many more. The distinction with the order on the reals (or subsets of the reals) is that the order relation is compatible with addition and multiplication. You can't do that in the complex numbers. The simple proof is to ask whether $i$ is greater or less than $0$. In either case, $i^2=-1$ should be greater than zero.

## Best Answer

The number $i^{i}$ is real, which immediately implies the result you're interested in. "Proof" : the complex conjugate of a complex number is obtained by replacing every occurrence of $i$ with $-i$; so the complex conjugate of $i^{i}$ is $(-i)^{-i}=(-1/i)^{i}=i^{i}$; but the only way that complex conjugation can give back the same number is if the number is actually real.

Addendum: strictly speaking, raising imaginary numbers to imaginary powers is not well-defined. Consider the following. We have $i=e^{i\pi/2}=e^{5i\pi/2}.$ Hence $i^{i}=e^{-\pi/2}=e^{-5\pi/2},$ but these two real numbers are clearly different. This is a delicate issue, in that resolving this contradiction requires choosing a "branch" of the complex logarithm.