Is $\left((-1)^2\right)^\frac12 = (-1)^\left(2\cdot\frac12\right)$?

Question

I'm feeling confused. If I square 1 and -1, the answers should be equal:

$1^2 = (-1)^2$

Then I take both sides to the power of $\frac12$:

$\left(1^2\right)^\frac12 = \left((-1)^2\right)^\frac12$

This next step seems to make sense according to the simple arithmetic rule about multiplying exponents:

$1^\left(2\cdot\frac12\right) = (-1)^\left(2\cdot\frac12\right)$

And then comes the weirdness:

$1^\left(\frac22\right) = (-1)^\left(\frac22\right)$

$1^1 = (-1)^1$

$1 = -1$

Obviously I did something wrong... Every step seems perfectly reasonable except going from step 2 to step 3. That seems reasonable too, that's what I was taught about exponents, but that's the only step which I could conceive has special constraints I violated. Is $\left((-1)^2\right)^\frac12 = (-1)^\left(2\cdot\frac12\right)$?

Answer 1

Note that $\sqrt{1} = \pm 1$. By squaring, you end up with extraneous solutions.

Another example is trying to solve $x=2$. By squaring both sides, we get $x^2 = 4$, which has solutions $x=2$ and $x-2$, hence $-2=2$, which is nonsense.