Is there a sense in which $4^{2} = -16$? Here's my train of thought:
I. $4^2=4^{(4/2)}$
II. $4^{(4/2)}=(4^4)^{(1/2)}$
III. $(4^4)^{(1/2)}=256^{(1/2)}$
IV. $256^{(1/2)}=\sqrt{256}$
square root of 256 has 2 solutions: 16 and -16
Can you explain me where I made the mistake and why am I not allowed to do whatever I did?
I thought that maybe (II) is wrong and only $4^{(4/2)}=(4^{(1/2)})^4$ is true,
but if $(x^a)^b=x^{(a\cdot b)}$ and numbers can switch places during multiplication and this does not change the outcome then:
$x^{(a\cdot b)}=x^{(b\cdot a)}$, and $(x^b)^a=x^{(b\cdot a)}$
Also, please don't insult me in the replies. I feel stupid enough already for not getting this.
 A: OVERALL THEORY
If $x$ is a positive real number and $m$ and $n$ are integers, then
$$x^{\frac mn} = (x^m)^{\frac 1n} = (x^\frac 1n)^m = \sqrt[n]{x^m} = (\sqrt[n]x)^m$$
When $x$ is negative, all heck breaks loose. We can say that
$$\text{$x^m$ is positive when $m$ is even and is negative when $m$ is odd.}$$
So there is no sense in which $4^2 = -16$.
It is OK to work with negative numbers and integer powers. For example,
$$((-2)^3)^4 = (-2)^{12} = 4096$$
But, weird things can happen with a negative $x$ and fractional exponents. For example
$$
\begin{array}{c}
   (-8)^\frac 13 = \sqrt[3]{-8}= -2 \\
   (-8)^\frac 26 = \sqrt[6]{((-8)^2)}= \sqrt[6]{64} = 2 \\
\end{array}
$$
Since $\frac 13 = \frac 26$, we see that we have a problem.
There is no way to fix this problem. There is no way around it. Raising a negative real number to a fractional power is not well-defined.
PRINCIPAL SQUARE ROOTS
The roots of an expression, $f(x)-n$, is the set of all $x$ for which $f(x)=0$
The solutions to the equation $x^2 - n = 0$ are called the square roots of $n$.

*

*Negative numbers have no (real-valued) square roots.

*$0$ has one square root.

*Positive numbers have two square roots.

The principal square root of a number is defined to be the non negative root of that number. There is a really cool equation that encapsulates this.
$$\sqrt{x^2} = |x|$$
Hence, for example,
$$\sqrt{(-3)^2} = \sqrt{3^2} = |3| = |-3| = 3$$
So, if $x^2 = y$, then the principal value of $\sqrt y$ is $|x|$.
ORIGIN OF THE WORD ROOT
A: Your I to IV are all correct. Here is what is incorrect: "square root of $256$ has two solutions." That is not true.
It is true that the equation $x^2 = 256$ has two solutions. But by convention, the square root  (written as $\sqrt{256}$ or $\sqrt{256}$ or $256^{1/2}$) is defined to be the non negative solution. It would have been possible to define it to be the negative one, but nobody does that. But it can never be both at the same time. A function like "sqrt" always needs to produce a single result.
