How the time complexity of this code is loglogn?

For n = k¹⁶:

iteration 1: i = k
iteration 2: i = k^2
iteration 3: i = k^4
iteration 4: i = k^16

So basically you can show that you are iterating log(m) times for k^m, where m = log(k).

Let us define the start value for i as j, then at the start of the first iteration, the value for i is j. At the end of the iteration, the value for i is j².

Now the next iteration, at the end, the value for i is j⁴. If at the start of the k-th iteration, the value is j^l, then at the end of the iteration, the value of j^2×l.

So that means at the k-th iteration, the value will be j^2k. We will keep iterating until we reach a value for i such that i > j, so that means that:

j^2k > n

we can here apply the log_j to both parts:

2^k > log_jn

and the log₂ to both:

k > log₂(log_jn)

for the first k for which this happens, the loop will end. So that means that this will happen "between" the log₂(log_jn) and log₂(log_jn)+1 iteration, but the +1 does not matter here.

The time complexity is thus: O(log₂(log_jn)) with j the initial value for i, and n the bound in the while loop.