Prove that $(1007!)^2 > \binom{2015}{1007}^2$ Prove that $(1007!)^2 &gt; \binom{2015}{1007}^2$.


This is equivalent to proving that $1007! &gt; \binom{2015}{1007}$. We then get this is equivalent to $$(1007!)^2 &gt; \dfrac{2015!}{1008!} = 2015 \cdot 2014 \cdots 1009.$$ How do we show this?
Consider the ratios $$A_n=\frac{(2n+1)!}{n!^2(n+1)!}$$ The goal is to show that $$A_{1007}&lt;1$$ To do so, compute $$A_8=\frac{2431}{4032}&lt;1$$ and note that $$\frac{A_{n+1}}{A_n}=\frac{2(2n+3)}{(n+1)(n+2)}&lt;1$$ for every $n\geqslant3$ since the difference of the denominator and the numerator is $$n^2-n-4\geqslant(n-3)(n+2)$$ This shows that $$A_n&lt;1$$ for every $n\geqslant8$.
You get $(1007!)^2$ compared to $\frac{2015!}{1008!}$.  The second expression is a product of $1007$ integers.  The first expression is the product of $2014$ integers.  If you take them in carefully chosen pairs, you can have the product of each pair greater than each factor in the second expression.