Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents?

I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.

But since $\sqrt x = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?

Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

How do I reconcile all this? What rules am I not aware of?

Edit: Since someone voted to close my question, I should probably explain the difference between my question and Proving square root of a square is the same as absolute value, regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there's any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I'm asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)

The rule $(x^a)^b = x^{ab}$ is only true for positive values of $x$. With negative values, you need to be much more careful.

For example, $\sqrt x \sqrt x = \sqrt{x\cdot x}$ is only true for positive values of $x$, because for negative values, the left side is not defined.

A common source of confusion or "paradoxes" comes from not paying close attention to the (perhaps rarely exercised) restrictions or boundary conditions. These restrictions are necessary to ensure that paradoxes like you're considering do not arise (i.e., otherwise the definitions would fail to be well-defined). For example, here's a proper definition of rational exponents from Michael Sullivan's College Algebra:

Note that we only consider real numbers here. Now, to answer your questions:

But since $\sqrt{x} = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?

The first assertion is not generally true; $\sqrt{x} = x^{\frac{1}{2}}$ only provided that $\sqrt{x}$ exists (that is, not for negative $x$). In your chained equality, the second equality is false, because the exponent in $x^{\frac{2}{2}}$ contains common factors (i.e., is not in lowest terms). These statements would, however, be true if $x$ was restricted to positive real numbers only.

Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

All of these equalities are false for negative $x$, because in that case the expression $\sqrt{x}$ does not exist in real numbers (i.e., it's undefined). Likewise, if you look carefully at the rule for multiplying radicals, then you'll see the same restriction against square roots of negative numbers.

Edit: Added text from Precalculus: a right triangle approach by Ratti & McWaters. Hopefully this clarifies the rule for products of square roots (namely that only positive radicands can be generally combined or separated). Also, note the warning from the section on complex numbers that doing so in that case is illegitimate.