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.)

First of all, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only when $a$ and $b$ are non-negative. $\sqrt{a}$ and $a^{\frac{1}{2}}$ are not the same function, as Cameron Williams points out in the comments above. Secondly, we can break your square of square root up. For positive values, $\sqrt{x^2} = x$, as one would expect. However, since $\sqrt{(-x)^2} = \sqrt{x^2}$, we notice that we get a positive answer from a negative value, getting $x$ from $-x$. We can represent this as flipping the function over the $x$-axis, or in other-words throwing a negative sign out front (For example, $y=-(x^2+5x)$ flips $y=x^2+5x$ across the $x$-axis). Thus, we get
$$\sqrt{\gamma^2}=\begin{cases}\gamma & \gamma\geq 0\\-\gamma & \gamma<0,\end{cases}$$
And this is the definition of the absolute value function
