Why is $\log(\log^*(n)) = Ω(\log^*\log(n))$ is true while $\log^*\log n$ is bigger than $\log \log^*n$?

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In field of algorithm and data structure I came across this problem. As far as I know

$$\lg^* \lg n > \lg \lg^* n$$

is true in term of growth rate and also noticed that

$$\log(\log^*(n)) = Ω(\log^*\log(n))$$

is True which means that $\lim f(n)/g(n)$ where $n$ goes to infinity in this case is infinity, meaning that growth rate of $\log(\log^*(n))$ is faster than $(\log^*\log(n))$ that is completely opposite to the previous term. Can someone explain?

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Note $\log^*(\log n)=(\log^* n)-1$ so you're comparing $(\log^*n)-1$ to $\log(\log^*n)$. Writing $\ell=\log^*n$, this is comparing $\ell-1$ to $\log \ell$; obviously $\ell-1$ will dominate $\log\ell$, so your second equality (asymptotic) is backwards, it should be $\log^*\log n=\Omega(\log\log^*n)$, which says $\log^*\log n$ is bounded below by / not dominated by $\log\log^* n$.