$n!\leq (n/\sqrt{2})^n$ for any positive integer $n>1$? Let $n$ be a positive integer such that $n>1$. Then, how do I prove that $n!\leq (n/\sqrt{2})^n$?
Let $K$ be a number field and $s$ be the number of non-real embeddings $K\rightarrow \mathbb{C}$.
Then, there exists a constant $M_K$ such that for any nonzero ideal $I$ of $O_K$, there exists a nonzero element $a\in I$ such that $|N_{K/\mathbb{Q}}(a)|\leq M_K \sqrt{|\Delta_K|}N(I)$.
$M_K$ can be taken to be $(\frac{4}{\pi})^s \frac{n!}{n^n}$, and this number is called the Minkowski bound.
However, Neukirsch, in his text, proves the above theorem with $M_K=(\frac{2}{\pi})^s$.
I'm not sure which one is smaller. Note that Neukirsch's $M_K$ is greater than the Minkowski's $M_K$ if we can prove that $n!\leq (n/\sqrt{2})^n$ for any positive integer $n>1$, but I am not sure how to prove this.
 A: The case $n=2$ can be checked separately. For $n\ge3,$ by the AM-GM inequality we have
$$(n!)^{\frac1n}\le\frac{1+2+\cdots+n}n=\frac{n+1}2\lt\frac n{\sqrt2},$$
so
$$n!\lt\left(\frac n{\sqrt2}\right)^n.$$
A: Wikipedia gives $ n! < (n/2)^n$ for $n\ge 6$, which implies $n!\leq (n/\sqrt{2})^n$.
The cases $2 \le n \le 5$ can be done manually.
Using the better estimate
$$
n! \leq e\left(\frac{n+1}e\right)^{n+1}
$$
it suffices to do the cases $2 \le n \le 3$ manually because
$$
e\left(\frac{n+1}e\right)^{n+1} \le \left(\frac{n}{\sqrt{2}}\right)^n
$$
for $n \ge 4$.
A: It is very easy to see by looking at $(n!)^2$ and pairing small terms with large terms, that  $$n! \le (\tfrac{n+1}{2})^n$$
for all $n\ge 1$.  We then easily verify that $(n+1)/2 < n/\sqrt2$ for all $n \ge 3$, leaving only the case $n=2$ to check separately.
A: A most elementary approach: Note that if $n \geq 2$ is even then \begin{align*} n! & = \bigg(1 \cdots \frac{n}{2}\bigg)\bigg(\big(\frac{n}{2}+1\big)  \cdots n\bigg) \\ & \leq \bigg( \frac{n}{2} \cdots \frac{n}{2} \bigg) \bigg( n \cdots n \bigg) \\ &= n^n / 2^{n/2} \end{align*}
If $n>2$ is odd then $n+1<\sqrt 2 n$, so similarly we get \begin{align*} n! &=  \bigg(1 \cdots \frac{n-1}{2}\bigg)\cdot \frac{n+1}{2}\cdot \bigg(\big(\frac{n+1}{2}+1\big)  \cdots n\bigg) \\ & \leq \bigg( \frac{n}{2} \cdots \frac{n}{2} \bigg)\cdot \frac{n}{\sqrt 2} \cdot \bigg( n \cdots n \bigg) \\ &= n^n / 2^{n/2} \end{align*}
