Numbers for which $n!=n^n$ I recently came across this relation valid for all positive integers ...which is $n!\leq n^n$ while it's proof is very easy by basic induction ...but I wanted to know for what values of n .. the equality($=$) holds..? I tried putting some values but I found only n=1 is satisfying it...are there any other values also..? if yes what are they? How can I find them without trial and error ...and ..if there are no other such values of n for which $n!=n^n$ how can one be so sure of that?
 A: Note that the equality holds only for $n=1$. Indeed you can see that $n^n=n\cdot n\cdot .....\cdot n$( $n$ times. While $n!=n(n-1)\cdot ....\cdot 1$
A: For $n>1$ we have
\begin{align}
n^n &= n! \iff \\
n \ln n &= \sum_{k=1}^n \ln k \iff \\
(n-1) \ln n 
&= \sum_{k=1}^{n-1} \ln k
< \sum_{k=1}^{n-1} \ln n = (n-1) \ln n
\end{align}
where for the last step we subtracted the last summand $\ln n$ from each side of the equation and for the comparison used that $\ln x$ is strictly increasing, thus $\ln k < \ln n$ for $k\in\{1,\ldots,n-1\}$.
$x < x$ is a false statement.
A: As an example, is $100!$ equal to $100^{100}$? Well:
\begin{align}
100!&=\phantom{00}1\cdot\phantom{10}2\cdot\phantom{10}3\dotsb\phantom199\cdot100\\
100^{100}&=100\cdot100\cdot100\dotsb100\cdot100
\end{align}
Since all but one of the factors of $100!$ are smaller than those of $100^{100}$, we have $100!<100^{100}$. This same argument shows that $n!<n^n$ for all $n>1$.
A: Obviously, except for $n=1$,$$\frac{n!}{n^n}=\frac1n\frac2n\frac3n\cdots\frac nn<1.$$
A: $n! = 1*2*....*n \le n*n*....*n = n^n$ with equality holding if and only if $i = n$ for all $i \le n$.  Well....
As $n - 1 < n, n -1 \ne n,$ the only natural number $n$ where $i \le n \implies i = n$ are where $n-1$ is not a natural number.  In other words $n = 1$.
A: $n! = \prod\limits_{i=1}^n i \lneq \prod\limits_{i=1}^n n = n^n$ for all $n>1$ since $i<n$ for all $i\in\{1,\dots,n-1\}$
