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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?

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    $\begingroup$ Try starting at $n = 1$ and see what happens. It should be hard to convince yourself there is an $N > 0$ such that $$n > N \ \Rightarrow \ n^n > n!$$ $\endgroup$ – Simon S Nov 8 '15 at 17:21
  • $\begingroup$ @Simon I honestly don't get what you are trying to say!! Can you please expand that a little bit??.. $\endgroup$ – Freelancer Nov 8 '15 at 17:27
  • $\begingroup$ It might be worth mentioning that $0!=1$ and in several (but not all) contexts, we allow $0^0$ to take the value of $1$ as well. If you include $0$ then, there are two solutions to $n!=n^n$ $\endgroup$ – JMoravitz Nov 8 '15 at 17:30
  • $\begingroup$ Look carefully at JMoravitz's answer below $\endgroup$ – Simon S Nov 8 '15 at 17:30
  • $\begingroup$ For a sharper inequality, see math.stackexchange.com/questions/1519559/… $\endgroup$ – Simon S Nov 8 '15 at 20:49
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$n! = \prod\limits_{i=1}^n i \lneq \prod\limits_{i=1}^n n = n^n$ for all $n>1$ since $i<n$ for all $i\in\{1,\dots,n-1\}$

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    $\begingroup$ @freelancer : more informally, $2! = 2 \cdot 1 < 2 \cdot 2 = 2^2$, $3! = 3 \cdot 2 \cdot 1 < 3 \cdot 3 \cdot 3 = 3^3$, $4! = 4\cdot 3 \cdot 2 \cdot 1 < 4^4$, etc. Hence in my notation above $$n > N = 1 \ \Rightarrow \ n^n > n!$$ $\endgroup$ – Simon S Nov 8 '15 at 17:33
  • $\begingroup$ @mvw i use it on occasion to emphasize the not equal part, but of course $\lneq$ has the same meaning as $\lt$. $\endgroup$ – JMoravitz Nov 8 '15 at 18:18
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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 $n!=n(n-1)\cdot ....\cdot 1$

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  • $\begingroup$ It also holds for $n=0$… ;-) $\endgroup$ – egreg Nov 8 '15 at 17:30
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    $\begingroup$ @egreg Oof, don't go into this $0^0$ nonsense, no good can come of it… :P $\endgroup$ – Akiva Weinberger Nov 8 '15 at 17:31
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    $\begingroup$ Just plot $\Gamma(x+1)$ against $x^x$. $\endgroup$ – vadim123 Nov 8 '15 at 17:37
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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.

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    $\begingroup$ This really looks good but can you explain a bit more on how you got this... Third step. ..(n-1)ln(n).. $\endgroup$ – Freelancer Nov 9 '15 at 2:39
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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 $n>1$.

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Obviously, except for $n=1$,$$\frac{n!}{n^n}=\frac1n\frac2n\frac3n\cdots\frac nn<1.$$

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$n! = 1*2*....*n \le n*n*....*n = n^n$ 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 $n = 1$.

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  • $\begingroup$ Oh, I guess I need to make a case for n= 0. But as $0^0$ is either 0, 1, or undefined (depending on whom you talk to). $0! = 1 \le 0^0$ $\endgroup$ – fleablood Nov 9 '15 at 2:52

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