Does $x^{n/n} = |x|$? Just a couple of small technical point here.  If x and n are real numbers, do we have to write $x ^ {n/n} = |x|$?  Or can we just reduce it to $x^{n/n} = x$?
One reason I ask is because then we would arrive at $x = x^1 = x ^{n/n} = |x|$.  Does this mean, then, that $x$ is not equal to $x^{n/n}$?  Or that $x ^ 1$ is not equal to  $x^{n/n}$?
Similarly, if I write $(\sqrt x)^2 = x^{2/2} = \sqrt{x^2}$, am I correct?  Or does the order matter here?
 A: You are incorrect. 
Once you are dealing with non-integer exponents, you either only define $x^y$ for $x$ positive, or you no longer have the property that $(x^y)^z = x^{yz}$ in general. 
The expression $x^y$ gets complicated when $y$ is rational, it gets stranger when $y$ is irrational, and it gets insane when $y$ is a complex number.  :)
Finally, the expression $x^{n/n}= x^1$ for any $x$, since $n/n=1$, so we have the order of operations - $n/n$ isn't some representation that is "like" $1$, it is $1$.
If you define $x^1=|x|$, you are losing a lot of rich mathematics. :)
A: You seem to be confusing $x^{n/n}$ with $(x^n)^{1/n}$. These are not quite equivalent (nor are any of them equivalent to $(x^{1/n})^n$. As others have pointed out, $x^{n/n}=x^1=x$.
More interestingly, consider $(x^n)^{1/n}=\sqrt[n]{x^n}$. Note that if $x\in \Bbb R$ and $n$ is a positive integer, then:

$$ \sqrt[n]{x^n}=\begin{cases}
x & \text{if $n$ is odd} \\
|x| & \text{if $n$ is even} \\
\end{cases}$$

A: if we interpret $x^{n/n}$ as $(x^{n})^{1/n}$ or $(x^{1/n})^{n}$, they are different: 
The first one contains a multivalued function
$$
f(z) = z^{1/q}
$$
so that $(x^{n})^{1/n} = \{xe^{2\pi i/n}, i = 0,1,\ldots,n-1\}$
The second one can be simplified to just a single value $x$, since suppose $x = re^{i\theta}$, then
$$
x^{1/n} = \{r e^{(2\pi i + \theta)/n}, i = 0,1,\ldots,n-1\}
$$
Then the mapping $z \mapsto z^n$ merges then together to a single $x$ again.
