I tried to solve this recurrence relation, like the following:
$$ \\ T(n) = \begin{cases} T(\sqrt[2]{n}) + c, & \text{if }n \geq 2 \\ a, & \text{otherwise} \end{cases} \\\\ T(n) = T(\sqrt[2]{n}) + c \\ T(n) = T(\sqrt[4]{n}) + c + c = T(\sqrt[2^2]{n}) + 2c \\ T(n) = T(\sqrt[8]{n}) + c + c + c = T(\sqrt[2^3]{n}) + 3c \\ ... \\ T(n) = T(\sqrt[2^k]{n}) + kc = T(n^{\frac{1}{2^k}}) + kc $$
The predicament is at the base case solving:
Using $n^{\frac{1}{2^k}} = 1$, it gets nowhere. Also, using $n^{\frac{1}{2^k}} = 2$ doesn't yield any interesting outcome.