TITLE: Prove that there are no positive integers $n>1$ and $k>1$ such that $n!=k^k$.
QUESTION [6 upvotes]: Prove that there are no positive integers $n>1$ and $k>1$ such that $n!=k^k$.

Can anybody help me?
if $n=k$ then $n!\neq k^k$,
if $n<k$, then $n!\neq k^k$,
if $n\geqslant2k$, then $n!\neq k^k$,
if $k<n<2k$ i don't know what to do. Maybe it is wrong way.

REPLY [8 votes]: Assume $n!=k^k$ with $n>1$.
Let $p$ be a prime $\le n$. Then $p\mid n!$ implies $p\mid k^k$, hence $p\mid k$.
So let $p_1=2,p_2=3,\ldots, p_r$ be all the primes $\le n$.
Then from the above, we have $p_1p_2\cdots p_r\mid k$ and in particular $k\ge p_1p_2\cdots p_r$.
From Euklid's proof of the infinitude of primes, we remember that there exists a prime $q\le p_1p_2\cdots p_r+1$ that differs from all $p_i$. Thus in our situation we conclude $n<q$ and in particular $n\le p_1p_2\cdots p_r\le k$.
But then $k^k\ge n^n>n!$, contradiction.