 # Comparing 2013! and 1007^{2013}. frobirrimupyx 2022-09-13 Answered
Comparing 2013! and ${1007}^{2013}$
I have to compare the following two numbers:

where $n!=1×2×\cdots ×\left(n-1\right)×n$
I tried in different ways to group the $1×2×\cdots ×2012×2013$ 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?
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Step 1
Use Stirling’s approximation for the factorial:
$n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^{n}\phantom{\rule{thickmathspace}{0ex}},$
so $2013!\approx 112.46356\left(740.54132{\right)}^{2013}<{1007}^{2013}\phantom{\rule{thickmathspace}{0ex}}.$
Step 2
For a less computational approach, look at their ratio:

The 1006 factors on the left are between 1 and 2; the factors on the right are between 1 and $\frac{1}{1007}$. Try matching them up.

We have step-by-step solutions for your answer! rustenig
Step 1
$1×2×\dots ×2012×2013$
$=\left(1007-1006\right)×\left(1007-1005\right)×\dots ×\left(1007+1005\right)×\left(1007+1006\right)$
$=\left({1007}^{2}-{1006}^{2}\right)×\left({1007}^{2}-{1005}^{2}\right)×\dots ×\left({1007}^{2}-{1}^{2}\right)×1007$
Step 2
$<\left({1007}^{2}{\right)}^{1006}×1007$
$={1007}^{2013}$

We have step-by-step solutions for your answer!