Iterated logarithm $\log^* n$
Let's assume the definition under which $\log^* 2 = 1$, $\log^* 2^2 = 2$, $\log^* 2^{2^2} = 3$, and so on. In order to show that $\log^* n = \omega(1)$, all you have to do is show that $\log^* n \to \infty$. Since $\log^*$ is increasing, it is enough to give a sequence of values $n_1,n_2,\ldots$ such that $\log^* n_i \to \infty$. Can you think of such a sequence?

In order to show that $\log^* n = o(\log^k n)$ for any fixed $k &gt; 0$, find numbers $n_\ell$ such that $n \leq n_\ell$ implies $\log^* n \leq \ell$, and notice that for $n_{\ell-1} &lt; n \leq n_\ell$, $\log^* n/ \log^k n \leq \ell/(\log^k n_{\ell-1})$. Perhaps you can show that $\log^k n_{\ell-1}$ is much larger than $\ell$ for large enough $\ell$. More concretely, you can try to show that $\log^k n_{\ell-1} \geq \ell^2$ (for example) for large enough $\ell$.