How to mathematically prove that a relation T(n)=T($\sqrt{n}$)+c is O(log(log(n))? following question, I understood the intuition behind how cutting down the size of input by square root on each iteration leads to O(log(log(n))) complexity.

I tried to derive it on paper.

Let T(n) = T($\sqrt{n}$) + c
$\implies$ T($n^{1/2}$)+2c
$\implies$ T($n^{1/4}$)+3c
$\implies$ T($n^{1/8}$)+4c
$\implies$ T($n^{1/16}$)+5c
.....upto n > 1

I noticed that the power of n becomes 1/2 times the last value.

How should I proceed from here on?
I need to derive that T(n) is O(log(log(n))).
Let $x = \log n$ and $Q(x) = T(2^x)$. You can rewrite your recurrence as follows:
$$
Q(x) = T(2^x) = T(n) = T(n^\frac{1}{2}) + c = T(2^{x/2}) + c = Q(x/2) + c.
$$

Which is easily solved using, e.g., the Master Theorem to obtain $Q(x) = \Theta(\log x)$. Substituting back:
$$
T(n) = Q(x) = \Theta(\log x) = \Theta(\log \log n).
$$