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Assume $n>1$. The solution to $k^k=n$ is $k=e^{W(\log n)}$, where $W(\cdot)$ is the Lambert W function. Another way to express the solution is $$k = \dfrac{\log n}{\log\dfrac{\log n}{\log\dfrac{\log n}{\log\dfrac{\log n}{\cdots}}}}.$$ In other words, $k$ is the limit of $k_0=\log n$, $k_1=\dfrac{\log n}{\log k_1}$, $\cdots$, $k_{t+1}=\dfrac{\log n}{\... 2 If a binary tree has height$h$then it has at most 1 node at depth 0, at most 2 nodes at depth 1, ..., at most$2^{h-1}$nodes at depth$h-1$(the maximal depth), and so at most$1+2+\cdots+2^{h-1} = 2^h-1$nodes in total. You can also prove this by induction. When$h=1$, the tree consists only of a root, so at most$1=2^h-1\$ vertices. Given a tree of ...