is there any difference between these three logarithms? is there any difference between $\log(n)^{\log(n)}$ vs $(\log n)^{\log n}$ vs $\log n^{\log n} $ from asymptotic growth rate? maybe all the same. I doubt about notation.
maybe this is very basic or crazy question, but I need to know.
 A: One must be careful, there is a crucial difference between $$(\log n)^{\log n}$$ and $$\log n^{\log n}$$ The exponent rule for the logarithm can be applied to this last one to obtain: $$\log n^{\log n}= \log n \log n=(\log n)^2 \neq (\log n)^{\log n}$$
A: Yes there is. Simple thumb rule: Power on the operand comes to side.
Suppose there is a logarithm exprerssion $\log_a{x}, a\neq1, x\gt0$. I am going to refer $x$ as the operand.Now, if an exponent $y$ is present on the operand $x$, then:
$$\log_a{(x^{y})} = y\log_a{x}$$
Proof:
Here I am considering $y\ge0$ (since I am assuming growth rates require only positive integral domain). A similar proof is present for negative y too.
$$\log_a{(x^{y})} = \log_a{(x\times x \times x\times x...y\ times)} = \log_a{x}+\log_a{x}+\log_a{x}...y \ times = y\log_a{x}$$
The other one is simply
$$(\log_a{x})^y = \log_a{x}\times\log_a{x}\times\log_a{x}...y \ times$$
EDIT:
$\log{(n)}^{\log{(n)}}$ is a little ambiguous. If up to me, I would consider it as an exponent on the operand (just because parenthesis are there about n)
EDIT 2:
To put it simply, one growth is polynomial$\times$logarithmic and other logarithmic exponentiated to logarithmic. To find the actual difference, you could find the ratio of their limits as n$\to\infty$. Then you can compare their relative growth rates.
