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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 imagine you can work out the answer. The key is Stirling's approximation If you do work it out, you should consider posting the answer yourself. (Yes, that is not only allowed, it is encouraged.) –  André Nicolas Nov 5 '11 at 8:23
    
It's true; it's a consequence of the fact that $n!^2\ge n^n$. As for how that's proved, perhaps I'll leave that unstated for now... –  Harry Altman Nov 5 '11 at 8:24
    
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?? –  nick Nov 5 '11 at 9:37
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