Proof for positive integer Prove that for any positive integers $m$ and $n$,  there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$ where $d$ is some integer in $\mathbb{N}$.

Hint 
Let $p_1,p_2,..,p_n$ be pairwise distinct primes. By the Chinese Remainder Theorem, the system
$$x \equiv 0 \pmod{p_1^m} \\
x+1 \equiv 0 \pmod{p_2^m} \\
.... \\
x+n-1 \equiv 0 \pmod{p_n^m}$$
has solutions.
