How to show that $n^{\ln(\ln(n))} = \ln(n)^{\ln(n)}$ I have verified that $n^{\ln(\ln(n))} = \ln(n)^{\ln(n)}$ by plugging in values for $n$, but do not understand why it is true. I am not aware of any $\log$ rules that can be used to simplify $n^{\ln(\ln(n))}$.
 A: Note that $n=e^{\ln n}$ so
$$n^{\ln (\ln n)}=e^{\ln n\cdot \ln(\ln n)}=(\ln n)^{\ln n}$$
A: Notice that if we can prove
$$a^{\log_b{c}}=c^{\log_b{a}}$$
then we have proved your problem with $a=n$, $b=e$, and $c=\ln{n}$.
We now do just that. Let $a, b, c$ be arbitrary real numbers. Notice that $b^{\log_b{a}}=a$. Also notice that $(x^y)^z=x^{yz}$ for real numbers $x,y,z$. With these two facts in hand we have that
\begin{align}
  a^{\log_b{c}} &=
  (b^{\log_b{a}})^{\log_b{c}} \\
  &= b^{(\log_b{a})(\log_b{c})} \\
  &= b^{(\log_b{c})(\log_b{a})} \\
  &= (b^{\log_b{c}})^{\log_b{a}} \\
  &= c^{\log_b{a}}
\end{align}
And so we are done.
A: Following Cameron Williams' hint, we take the logarithm of both sides to get
$$\ln \left(n^{\ln\left(\ln (n)\right)}\right) = \ln \left( \ln (n)^{\ln (n)}\right)$$
Recall that $\log_b a^c = c\log_b a$. Use this property on both sides:
$$\left(\ln\left(\ln(n)\right)\right)\left(\ln(n)\right)=\left(\ln(n)\right)\left(\ln\left(\ln(n)\right)\right)$$
This is true because multiplication is commutative.
