Is $n! \nmid n^n$ true for every sufficiently large $n$ (maybe $n \ge 3$)?

Question

Is $n! \nmid n^n$ true for every sufficiently large $n$ (maybe $n \ge 3$)?

If so, how to prove it?

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Note that it is easy to show that $n! \nmid n^n$ if $n$ is a prime.

Answer 1

If $n!|n^ n$ then $n-1$ divides $n^n$.

$n$ and $n-1$ are coprime, it follows $n-1$ and $n^n$ are coprime, it follows $n-1$ does not divide $n^n$ unless $n-1=1$ which is not true as $n\geq 3$

Answer 2

Bertrands postulate.

If $p_k$ is the largest prime less than or equal to $n$ then there is a prime $p_{k+1}$ between $p_k < p_{k+1}< 2p_k - 2$.

But if $n!|n^n$ then the second largest prime $p_{k-1}$ is also a factor. SO $n > p_{k-1}p_k > 2p_k -2 > p_{k+1}$.

So $p_{k+1}| n!$ but $p_{k+1}\not \mid n$.

A contradiction.