elementary number theory - 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!