Show $(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))}$ I am having difficulty understanding how this follows.
$$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))} = n^{\log \log n}$$
Which logarithmic identities are used to go through each equality?
e.g. how do you first go from 
$$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))}$$
and then to
$$2^{(\log n)(\log (\log n))} = n^{\log \log n}$$
(The log base must be 2 or else this equality won't hold)
 A: I'm assuming the base is $2.$ Otherwise the equality doesn't hold.
$$2^{(\log n)(\log (\log n))} = 2^{(\log (\log n)) (\log n)} = (2^{(\log (\log n))})^{(\log n)}$$
and
$$ 2^{(\log (\log n))} = \log n $$
because
$$ 2^{\log x} = x.$$

Edit: to reflect the update in the question:
$$ 2^{(\log n)(\log (\log n))} = (\color{blue}{2^{(\log n)}})^{(\log (\log n))} = \color{blue}{n}^{{(\log (\log n))}} = n^{\log \log n} $$
because again
$$ 2^{\log n} = n.$$
A: For any
$$1\neq a\in\Bbb R^+\,\,,\,\,x^y=a^{y\log_ax}\,\,,\,\text{whenever LHS is defined. }$$
Thus,
$$(\log n)^{\log n}=2^{\log n\,\log_2(\log n)}$$
So your equality follows if the logarithm here is taken in base $\,2\,$ and not $\,10\,$ , as you wrote...
A: Let
$$y=\log_2^{\log_2 n} n$$
Taking the logarithm (in base 2) of both sides
$$\log_2 y=\log_2 n \log_2 \log_2 n$$
Now, remember that $2^{\log_2 n}=n$.  Thus
$$y=2^{\log_2 y}=2^{\log_2 n \log_2 \log_2}$$
Also recall that $a^{bc}=(a^b)^c$.  Thus
$$y=2^{\log_2 n \log_2 \log_2 n}=(2^{\log_2 n})^{\log_2 \log_2 n}=n^{\log_2 \log_2 n}$$
