Comparing $2013!$ and $1007^{2013}$ I have to compare the following two numbers:
$$2013! \text{ and } 1007^{2013}$$
where $n! = 1 \times 2 \times \cdots \times (n-1) \times n$.
I tried in different ways to group the $1 \times 2 \times \cdots \times 2012 \times 2013$ to obtain some kind of association with the $1007$ from $1007^{2013}$ but no luck.
Is there any standard approach for this kind of problem? 
 A: Let $j \in \{1,2,\dots,2013\}$. 
When $j$ is small, $1007$ is much larger in size than $j$. [For e.g. $1007 = 1007 \cdot 1$] So for small $j$, the $j$-th factor in $1007^{2013}$ is much larger in size than the $j$-th factor in $2013!$.
But when $j$ is large, $j$ is not that much larger than $1007$. [For e.g. $2013 < 1007 \cdot 2$] So for large $j$, the $j$-th factor in $2013!$ is not that much larger in size than the $j$-th factor in $1007^{2013}$.
Does this line of reasoning help you find an answer to this question?
[Notice also that $1007$ is the median (midpoint) of $\{1,2,\dots,2013\}$.] 
A: $1\times 2\times\ldots\times 2012\times 2013$
$=(1007-1006)\times(1007-1005)\times\ldots\times (1007+1005)\times(1007+1006)$
$=(1007^2-1006^2)\times (1007^2-1005^2)\times\ldots\times (1007^2-1^2)\times 1007$
$<(1007^2)^{1006}\times 1007$
$=1007^{2013}$
A: Hint:     Use the inequality $k(n-k)\le (n/2)^2$. Then you could prove the more general inequality (without  Stirling).
A: $1007^{2013}=(2013-1006)(2012-1005)(2011-1004).....(1008-1)$
$2013!=(1007+1006).(1007+1005).(1007+1004)....(1007-1006).(1007)$
$(1007^2-1006^2).(1007^2-1005^2).(1007^2-1004^2)...(1007^2-1^2).1007<(1007^2)^{1006} .1007$
$<1007^{2013}$
A: You can just take the inequality between the arithmetic and geometric means:
$$\sqrt[n]{a_1a_2\dots a_n}\leq \frac{a_1+a_2\dots+a_n}{n}$$
with $a_i=i$ and $n=2013$. Then $\sqrt[2013]{2013!} \leq 1007$.
A: A mere generalization, using Stirling's approximation again:
$$
n! \ \text{vs} \bigg(\frac{n}{2} \bigg)^n\\
\bigg(\frac{n}{e}\bigg)^n \sqrt{2 \pi n} \ \text{vs} \ \bigg(\frac{n}{2} \bigg)^n\\
2 \pi n \ \text{vs} \bigg(\frac{e}{2} \bigg)^{2n}
$$
LHS is $f(n)=2 \pi n=O(n)$, RHS is $g(n)=(\frac{e}{2} )^{2n}=O(a^{2n}), \ a>1$, so $f(n)=o(g(n))$. In fact this holds already for $n=6$.
A: Use Stirling’s approximation for the factorial:
$$n!\sim\sqrt{2\pi n}\left(\frac{n}e\right)^n\;,$$
so $$2013!\approx 112.46356(740.54132)^{2013}<1007^{2013}\;.$$
For a less computational approach, look at their ratio:
$$\begin{align*}\frac{2013!}{{1007}^{2013}}&=\frac{2013\cdot2012\cdot\ldots\cdot2\cdot1}{1007\cdot1007\cdot\ldots\cdot1007\cdot1007}\\\\
&=\underbrace{\frac{2013}{1007}\cdot\frac{2012}{1007}\cdot\frac{2011}{1007}\cdot\ldots\cdot\frac{1008}{1007}}_{1006\text{ factors}}\cdot\frac{1007}{1007}\cdot\underbrace{\frac{1006}{1007}\ldots\cdot\frac2{1007}\cdot\frac1{1007}}_{1006\text{ factors}}\;.
\end{align*}$$
The $1006$ factors on the left are between $1$ and $2$; the factors on the right are between $1$ and $\frac1{1007}$. Try matching them up.
