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Can anyone explain how can i prove this?$$\left(\frac{n}{2}\right)^{\frac n2}\leq n! \leq n^n$$ It seems I should use Stirling’s approximation, but I'm not sure about details.

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    $\begingroup$ Basically : $$ \begin{align} 1 \cdot 1 \cdot ... &\cdot n/2 \cdot \ \ \ n/2 \cdot \ ... \cdot n/2 \\ \leq 1 \cdot 2 \cdot ... &\cdot n/2 \cdot (n/2 + 1) \cdot ... \cdot n \\ \leq n \cdot n \cdot ... &\cdot n \ \ \cdot\ \ \ n \ \ \ \cdot\ \ \ \ \ \ ... \cdot n \end{align} $$ No need of fancy Stirling here. $\endgroup$ – Zubzub Sep 26 '17 at 11:13
  • $\begingroup$ Yes, in words the above comment essentially means compare term by term. $\endgroup$ – Jihoon Kang Sep 26 '17 at 11:13
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One way is obvious : Since $n!$ is the product of all numbers (there are $n$ of them) less than or equal to $n$, and $n^n$ is the product of $n$ with itself $n$ times, it should be clear that $n! < n^n$.

For the other one, note that if $n = 2k$, then $n! = n \times 2(n-1) \times 3(n-2)... \times k(k+1)$.

Note that each term is greater than or equal to $k$ above, and there are $k$ such terms, so $n! \geq k^k$.

Suppose $n = 2k-1$, then $n! = n \times 2(n-1) \times 3(n-2) \times ... \times k^2$. Note that each term is greater than or equal to $k$, and there are $k$ such terms, so again $n! > k^k$. But $k > \frac n2$ ,so clearly the inequality follows.

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