Prove that $n!$ is not a divisor of $n^n$ for $n>2$ Proof by induction.
$3! = 6$ is not a divisor of $3^3 = 27$
Now suppose that
$n!$ is not a divisor of $n^n$
$(n+1)n!= (n + 1)!$ is not a divisor of $(n + 1)n^n = n^{n + 1} + n^n $
I don't know how to go on. Maybe induction is not the way to prove the statement
 A: Let $p\leq n$ be prime.  Then $p|n!$ so if $n!|n^n$ then $p|n^n\implies p|n.$  Since $n>2$, $n-1>1$ so there is a prime $p$ that divides $n-1$.  By the above, also $p|n$ so $p|(n-(n-1))$, contradiction.
A: We will prove there is a prime number in $\{1,2,...,n\}$ which does not divide $n$.
Let $p_i$ be the $i^{th}$ prime.
Suppose the primes in $\{1,2,...,n\}$ are $p_1,p_2,...,p_m$. Suppose all of them divide $n$. Because $p_1,p_2,...,p_m$ are the primes in $\{1,2,...,n\}$, $p_{m+1}>n$. But $p_1,p_2,...,p_m$ divide $n$, so $n\geq p_1\cdot p_2\cdot...\cdot p_m$
So we get:
$$p_{m+1}>p_1\cdot p_2\cdot...\cdot p_m>2p_m$$
However, from Bertrand's postulate, if $p_{m+1}>2p_m$, there must be a prime between $p_{m+1}$ and $p_m$, which is a contradiction, since $p_m$ and $p_{m+1}$ are consecutive primes. (red about Bertrand's postulate here)
To conclude, it is impossible for all primes in $\{1,2,...,n\}$ to divide $n$, so let $p$, $1\leq p\leq n$ such that $p$ does not divide $n$. Because $p$ divides $n!$, if $n!$ divided $n^n$, it means $p$ would divide $n^n$, contradiction.
A: Assume $k^k$ is not a multiple of $k!$
Let's show $(k+1)^{k+1}$ is not a multiple of $(k+1)!$ that is equivalent to $(k+1)^k$ is not a multiple of $k!$
Since $k^k$ is not a multiple of $k!$, $k$ and some factors of $k!$ are relatively prime.
Since $k$ and $k+1$ are relatively prime, $(k+1)^k$ is not divisible by $k!$.
