Given recurrence relation : $$ T(n) = \begin{cases} T(n-\log n) + 1 & \text{if } n \ge 1, \\ 1 & \text{otherwise.}\\ \end{cases} $$
To find asymptotic order of $T(n)$ i do as follow:
Suppose $n=\log m$ now :
$$T(\log m)=T(\log m-\log \log m) +1$$
$$\implies T(\log m)= T(\log \frac{m}{\log m}) +1$$
,but in this step i get stuck and i don't know how we can transfer relation to get easier relation.