Suppose $N$ is a large integer and let $k$ be an integer s.t. $k > N$. Under what conditions can we conclude that $1^N, 2^N, \ldots, k^N$ are all divisors of $k!$?
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Trivially, the result is true for $N=1$, so consider the case where $N\geq 2$.
Suppose $k$ is even, then by Bertrand's Postulate, there is a prime $p$ satisfying $k/2 < p < k$, so that $2p > k$, and it follows that $p$ is a factor of $k!$, but $p^2$ is not, since there is only one multiple of $p$ amongst the numbers $1, 2, \dots, k$.
A similar argument works if $k$ is odd, say $k=2m+1$, since there is then at least one prime satisfying $m+1 < p < 2m+2$.