Prove that there exist $135$ consecutive positive integers so that the $n$th least is divisible by a perfect $n$th power greater than $1$ Prove that there  exist  135  consecutive  positive  integers  so  that  the  second  least  is  divisible  by  a perfect square $> 1$, the third least is divisible by a perfect cube $> 1$, the fourth least is divisible by a perfect fourth power $> 1$, and so on.
How should I go about doing this? 
I thought perhaps I should use Fermat's little theorem, or its corollary? 
Thanks!

By the Chinese remainder theorem, there is an integer $n$ such that $n\equiv -k\ (\text{mod}\ p_k^{k+1})$ for all $k=1,2,\dots 134$,  (where $p_k$ is the $k^\text{th}$ smallest prime). Then $n, n+1, \dots, n+134$ satisfy the required condition. 
