Special cases of the Laws of Exponents? I have a few questions regarding powers. 
We have:
\begin{align}
(x\cdot y)^n &= x^n\cdot y^n \tag 1\\
(x^m)^n &= x^{m\cdot n} \tag 2\\
x^m \cdot x^n &= x^{m+n} \tag 3
\end{align}
Question 1:
I guess a special case of $(1)$ is $y=x$, so:
\begin{align}
(x\cdot x)^n
\underbrace{=}_\text{$(1)$}
x^n\cdot x^n 
\underbrace{=}_\text{$(3)$}
x^{n+n}
=x^{2n}
\tag 4
\end{align}
Is this correct?
Question 2:
If a special case of $(3)$ is $m=n$ we have:
$$
x^n\cdot x^n
\underbrace{=}_\text{$(3)$}
x^{n+n}=x^{2n} \tag 5
$$
Is this correct?
But if $m=n$ in $(2)$ we have:
$$(x^n)^n 
\underbrace{=}_\text{$(2)$}
x^{n\cdot n} = x^{n^2} \tag 6
$$
And this doesn't make sense...
I guess it contradicts $(4)$ and $(5)$?
Thanks in advance!
 A: I think your confusion may be the fact that you think you calculate the same expression three times, but in the first two cases you compute $\color{blue}{x^n \cdot x^n}$ in two ways (or using two properties) while in the last case, you compute $\color{purple}{(x^n)^n}$ and that's not the same:
$$\underbrace{\color{blue}{x^n \cdot x^n}}_{2\text{ factors}} = (x^n)^2 \color{red}{\;\mathbf{\ne}\;} \color{purple}{(x^n)^n} = \underbrace{x^n \cdot x^n\cdot \ldots\cdot x^n}_{n\text{ factors}} $$
A: From the looks of it, you expected $x^n\cdot x^n$ and $(x^n)^n$ to be the same, somehow?
Well, they are not. If we expand the exponents into repeated products of $x$ (this is assuming $n$ is a natrural number), we see that
$$
x^n\cdot x^n = \underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{n\text{ times}}
$$
for a total of $2n$ $x$'s, while
$$
(x^n)^n = (\underbrace{x\cdot x\cdots x}_{n\text{ times}})^n\\
= \underbrace{\underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdots\underbrace{x\cdot x\cdots x}_{n\text{ times}}}_{n\text{ times}}
$$
for a total of $n\cdot n$ $x$'s.
Of course, if $n = 2$, or $n = 0$, or $x = 1$ or $x = 0$, then they happen to be the same. But in general they are quite different.
A: Question 1
This is indeed correct. In fact, another way to see it is to work from your equation (2):
$$ (x \cdot x)^n = (x^2)^n \underbrace{=}_{(2)} x^{2n} $$
Question 2
This is also correct.
Question 3
Why does this not make sense? Consider a simple example with $x=2, m=2, n=3$. Equation (2) provides
$$(x^m)^n = x^{mn} = 2^{2 \cdot 3} = 2^6 = 64$$
If we work it out without using the identity, we obtain
$$(x^m)^n = (2^2)^3 = 4^3 = 64$$
which is the same.
If you think the identity does not make sense, then what would your expected result be? Try an example and see if it is correct.
I hope this helps!
