$$\ T(n) = \sqrt{n} \cdot T(\sqrt{n}) + 1 $$
I've found so many similar questions but I couldn't understand any of the answers explanations. When I try to draw a recurrence tree, I see that each 'level' has as many operations as nodes (because of only $\ 1 $ operation in each node) so in the first level it has $\ 1 $ node then $\ \sqrt{n} $ nodes then $\ \sqrt{\sqrt{n}} $ nodes and so forth to $\ n^{\frac{1}{2^k}} $ on the lowest level on the tree.
I get the same answer when unrolling it:
$\ T(n) = n^\frac{1}{2} T(n^{\frac{1}{2}}) + 1 = n^{\frac{1}{2}}(n^{\frac{1}{4}} T(n^{\frac{1}{4}}) + 1) + 1= n^{\frac{1}{2}} ( n^{\frac{1}{4}}(n^{\frac{1}{8}} T(n^{\frac{1}{8}}) + 1 ) + 1) + 1 = ... $
But it is really inconvenient working with this form.
Also tried using substitution as mentioned here and then applying master theorem but with I can't understand how to make the transition back. Also a similar question here but no further explanations in the answers. I would rather use tree recurrence to solve it but substitution and master theorem also good.