$\log^*(n)$ runtime analysis So I know that $\log^*$ means iterated logarithm, so $\log^*(3)$ = $(\log\log\log\log...)$ until $n \leq 1$.

I'm trying to solve the following:

is 


  $\log^*(2^{2^n})$


little $o$, little $\omega$, or $\Theta$ of


  ${\log^*(n)}^2$


In terms of the interior functions, $\log^*(2^{2^n})$ is much bigger than $\log^*(n)$, but squaring the $\log^*(n)$ is throwing me off. 

I know that $\log(n)^2$ is $O(n)$, but I don't think that property holds for the iterative logarithm.

I tried applying the master method, but I'm having trouble with the properties of a $\log^*(n)$ function. I tried setting n to be max (i.e. $n = 5$), but this didn't really simplify the problem.

Does anyone have any tips as to how I should approach this?
Recall that for $k &gt; 1$, by definition we have $\log^*k = \log^*(\log{k}) + 1$.

By applying the definition twice, we see that $\log^*(2^{2^n}) = \log^*n + 2$.  Now we can compare $\log^*n + 2$ and $(\log^*n)^2$.