Please refer this https://gateoverflow.in/1243/gate2007-45
$T(n) = T(\sqrt{n}) +1;$
Let $n = 2^{m} $
$T(2^{m}) = T(\sqrt{2^{m}}) +1;$
$T(2^{m}) = T(2^{m/2}) +1;$
Now let $T(2^{m}) = S(m)$
So $S(m) = S(\frac{m}{2} ) +1$
It is binary search recurrence relation
so $S(m) = O(log m)= O(log log n)$ ($\because n = 2^{m}$)
since $T(n) = T(2^{m}) = S(m)$
T(n) is $O(log log n)$