What is the time complexity for this Algorithm? -

for = 1 n     j = 2     while j <         j = j * j 

i think it's time complexity : log(n!) = n * log(n).

but solution said : n * loglog(n) , didn't understand why?

in explanation below, assume arithmetic , comparison operations o(1).

for = 1 n 

the below repeated n times, makes n * part in solution.

    j = 2     while j <         j = j * j 

the above calculates first number of following sequence that's >= i :

2 = 2^(2^0) 4 = 2^(2^1) 16 = 2^(2^2) 256 = 2^(2^3) 65536 = 2^(2^4) ... 

so thing need find relation between , 2^(2^i). , log(log(2^(2^i))) = log(2^i) = i.