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!

Use the Chinese Remainder Theorem.  Pick $134$ distinct primes.  The perfect square is the square of the first, the cube is the cube of the second, and so on.  All your moduli are distinct, so CRT guarantees a solution.  If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 \equiv 0 \pmod {2^2}, N+2 \equiv 0 \pmod {3^3}, N+3 \equiv 0 \pmod {5^4}\ldots$
