By backward substitution, We may finally get like this,
k T(n^(1/2^k))+k-1
and given that T(2)=1
So, n^(1/2^k)=2
1/2^k . logn = 1
2^k = logn
K.log2 = log(log n)
So, our equation will be,
log(logn)[1]+log(log n) -1
O(log log n)
I followed this same procedure even in GATE, but I did very minor mistake and I wrote equation like this, "k T(n^(1/k))+k-1 and my whole answer changed to log(n) only which is technically right for me, but question wise not! Such a big mistake!!!!