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If so, how can it be proven? (I have evaluated it up to $n=25$.)

If not, does there exist a $k\in\mathbb{R}$ such as that $n> a^k\Rightarrow n!>a^n$, with $n\in\mathbb{N},a\in\mathbb{R}$?


It is true, and to prove it, it suffices to show that $n!^2\geq n^n$ with induction.

For $n=1$ we have that $1\geq1$, which is true. Suppose $n!^2\geq n^n$. Then, $(n+1)!^2=(n+1)^2n!^2\geq (n+1)^2n^n$. We need to show that

$\begin{align} &(n+1)^2n^n\geq (n+1)^{n+1}\Leftrightarrow\\ \Leftrightarrow &n^n \geq (n+1)^{n-1}\Leftrightarrow\\ \Leftrightarrow &\ln(n^n)\geq \ln((n+1)^{n-1})\Leftrightarrow\\ \Leftrightarrow &n\ln n\geq (n-1)\ln(n+1)\Leftrightarrow\\ \Leftrightarrow &\frac{n}{n-1} \geq \ln(n+1-n)=\ln1=0\quad, \end{align}$

which is true!!

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    I proved $n!^2\geq n^n$ with induction.For $n=1$ we have that $1\geq1$ which is true. Suppose $n!^2\geq n^n$. Then, $(n+1)!^2=(n+1)^2n!^2\geq (n+1)^2n^n$. We need to show that $(n+1)^2n^n\geq (n+1)^{n+1}\Leftrightarrow (n+1)n^n\geq (n+1)^n\Leftrightarrow n^n \geq (n+1)^{n-1}$ $\Leftrightarrow \ln(n^n)\geq \ln((n+1)^{n-1})\Leftrightarrow n\ln n\geq \(n-1)\ln(n+1)$ $\Leftrightarrow \frac{n}{n-1} \geq \ln(n+1-n)=ln1=0$ which is true!! How could I use the Stirling's approximation to get a similar result??2011-11-05

1 Answers 1

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Converting above realization to an answer.


We prove $ \left( n! \right)^2 \ge n^n $ by means of induction.

For $n=1$, we have $ 1 \ge 1 $, which is true. Now suppose that $ \left( n! \right)^2 \ge n^n $. Then, $ \begin {align*} \left( (n+1)! \right)^2 &= \left( n + 1 \right)^2 \left( n! \right)^2 \\&\ge \left( n + 1 \right)^2 n^n. \end {align*} $Furthermore, $ \begin {eqnarray*} \left( n+1 \right)^2 n^n &\ge& (n+1)^{n+1} \\ \iff \left( n + 1 \right) n^n &\ge& (n+1)^n \\ \iff n^n &\ge& (n+1)^{n-1} \\ \iff \ln \left( n^n \right) &\ge& \ln \left( \left( n + 1 \right)^{n-1} \right) \\ \iff n \ln n &\ge& \left( n - 1 \right) \cdot \ln \left( n + 1 \right) \\ \iff \frac {n}{n-1} &\ge& \ln \left( n + 1 - n \right) = \ln 1 = 0, \end {eqnarray*} $ which is true, so we are done. $\Box$