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Note that for integers $i > 2$ and. $h > 0$, and except for the case $(i,h) = (3,1)$, one has $i^h(h!) < 2^{\lfloor ih/2 \rfloor}(\lfloor ih/2 \rfloor !)$. We can account for the exception and bound from above the denominator of the left hand side of the posted inequality by $(3)2^{(n - c_1)/2}(((n - c_1)/2)!)(c_1!) \leq (3)2^{(n - c_1)/2}(((n + c_1)/2)!)$. This latter term is increasing in $c_1$, and for large $n$ one can have $c_1 \leq n - 10$ and still be less than $(n - 4)!$. One can use this to show the inequality of the post is satisfied for $n > 15$ and $c_1 < 4$; likely the inequality holds for more $n$ and more $c_1$.
Letting $f(i)=i^{c_i}(c_i)!$, we can rearrange the poster's inequality to $$(n - 4)! \gt \prod_{1 \le i \le n}f(i)$$, and ask for which values of $n, i,$ and $c_i$ the inequality holds.  Given $c_i$, let $g(i) = 2^{\lfloor ic_i/2 \rfloor} (\lfloor ic_i /2 \rfloor)!$ for $i \gt 1$ and $g(1) = f(1)$.  Now, when $i = 3$ and $c_i = 1$ we have $f(3) \lt 2g(3)$, and for other pairs $(i,c_i)$ with $i \gt 2$ and $c_i \gt 0$ one has $f(i) \le g(i)$, so one can have the original inequality follow from $$(n-4)! \gt 2\prod_{1 \le i \le n} g(i)$$.   However, the product of the $g(i)$ is itself bounded from above by $h=2(c_1)! 2^{(n - c_1)/2} {\lceil(n - c_1)/2\rceil}!$.  When $c_1 \le 4$ and $n \ge 15$,  $h$ 2h$is less than$(n-4)!$. It is clear that the original inequality implies$c_1 \lt n-4$, and that there is no solution for$n \lt 8$. For$n=8$, the product of the$f(i)$has to be less than 24,$24$, so$c_1 \lt 4$. For$n= 8,9,10,11$it is routine to find restrictions on$c_1$that will permit solutions of the inequality. From another direction, if$c_1 = n-5$then case-by-case examination gives that$f(i) = 1$for$ i \gt 5$and$f(2)f(3)f(4)f(5)$is at most 8,$8$, so that$n > 12$for the original inequality to hold. When$c_1 = n - 6$, a similar analysis requires$n > 11$, and for larger values of$n - c_1$the inequality holds for all the remaining meaningful cases. END EDIT 2012.01.05 Gerhard "Ask Me About System Design" Paseman, 2011.12.26 3 added 1486 characters in body BEGIN EDIT 2012.01.05I decided to add some detail to the post. Letting$f(i)=i^{c_i}(c_i)!$, we can rearrange the poster's inequality to $$(n - 4)! \gt \prod_{1 \le i \le n}f(i)$$, and ask for which values of$n, i,$and$c_i$the inequality holds. Given$c_i$, let$g(i) = 2^{\lfloor ic_i/2 \rfloor} (\lfloor ic_i /2 \rfloor)! $for$i \gt 1$and$g(1) = f(1)$. Now, when$i = 3$and$c_i = 1$we have$f(3) \lt 2g(3)$, and for other pairs$(i,c_i)$with$i \gt 2$and$c_i \gt 0$one has$f(i) \le g(i)$, so one can have the original inequality follow from $$(n-4)! \gt 2\prod_{1 \le i \le n} g(i)$$. However, the product of the$g(i)$is itself bounded by$h=2(c_1)! 2^{(n - c_1)/2} {\lceil(n - c_1)/2\rceil}!$. When$c_1 \le 4$and$n \ge 15$,$h$is less than$(n-4)!$. It is clear that the original inequality implies$c_1 \lt n-4$, and that there is no solution for$n \lt 8$. For$n=8$, the product of the$f(i)$has to be less than 24, so$c_1 \lt 4$. For$n= 8,9,10,11$it is routine to find restrictions on$c_1$that will permit solutions of the inequality. From another direction, if$c_1 = n-5$then case-by-case examination gives that$f(i) = 1$for$ i \gt 5$and$f(2)f(3)f(4)f(5)$is at most 8, so that$n > 12$for the original inequality to hold. When$c_1 = n - 6$, a similar analysis requires$n > 11$, and for larger values of$n - c_1$the inequality holds for all the remaining meaningful cases. END EDIT 2012.01.05 2 added 27 characters in body Note that for integers$i > 2$and.$h > 0$, and except for the case$(i,h) = (3,1)$, one has$i^h(h!) < 2^{\floor{ih/2}}(\floor{ih/2}!)$. 2^{\lfloor ih/2 \rfloor}(\lfloor ih/2 \rfloor !)$. We can account for the exception and bound from above the denominator of the left hand side of the posted inequality by $(3)2^{(n - c_1)/2}(((n - c_1)/2)!)(c_1!) < <= \leq (3)2^{(n - c_1)/2}(((n + c_1)/2)!)$. This latter term is increasing in $c_1$, and for large $n$ one can have $c_1 <= \leq n - 10$ and still be less than $(n - 4)!$. One can use this to show the inequality of the post is satisfies satisfied for $n > 15$ and $c_1 < 4$; likely the inequality holds for more $n$ and more $c_1$.