Asymptotic $T(n)=T(\sqrt{n})+1$

Question

I would like to find the complexity of $T(n)=T(\sqrt{n})+1$

I did : $$T(n)=T(\sqrt{n})+1$$ $$T(n)=T(n^{1/2})+1$$ $$T(n)=(T(n^{1/4})+1)+1=T(n^{1/4})+2$$ And after $k$ steps : $$T(n)=T(n^{\frac{1}{2^k}})+k$$

How should I continue ?

Answer 1

Take $n=m^{2^k}$, then $$ T(m^{2^{k+1}})=T(m^{2^k})+1 $$ and if $S_m(k)=T(m^{2^k})$, then $S_m(k)=k+S_m(0)$.

Hence $$ T(n)=\log_2(\log n)+c. $$