Calculation of factorial numbers, and product in increasing order. Arrangement of following product in increasing order from left to right:
$A = 1000!$
$B = (400!)^2⋅(200!)$
$C = (500!)^2$
$D = (600!)⋅(300!)⋅(1000!)$
$E = (700!)⋅(300!)$
$\underline{\bf{My\; Try}}$: Here $(600!)\cdot(300!)\cdot(1000!)$ is largest, but I did not understand in which sequence the order is increase.
Help Required.
Thanks.
 A: A useful lemma that you should prove:

Lemma: $(n - p)!\ p!\ \leq\ n!$ with equality only when $p = 0$ or $n$.

If you are having trouble with the proof of this fact, I have provided a hint below (put your cursor over the grey box).

 Proof: Consider ${n \choose p}$.

This lemma tells us that $B, C, E < A$. Clearly $A < D$ so the chain ends with $A < D$. We just need to determine the relative sizes of $B, C,$ and $E$.
Note that $$\frac{B}{C} = \frac{(400!)^2 200!}{(500!)^2}=\frac{200!}{\left(\dfrac{500!}{400!}\right)^2}=\frac{200\times 199\times \dots\times 2\times 1}{(500\times 499\times\dots\times 402\times 401)^2} < 1$$ as each factor in the denominator is larger than each factor in the numerator, so $B < C$. 
Similarly, one can compare $B$ and $E$ by considering $$\frac{B}{E} = \frac{(400!)^2 200!}{700!\ 300!} = \frac{\left(\dfrac{400!}{300!}\right)200!}{\dfrac{700!}{400!}}.$$ Likewise for $C$ and $E$ $$\frac{C}{E} = \frac{(500!)^2}{700!\ 300!} = \frac{\dfrac{500!}{300!}}{\dfrac{700!}{500!}}.$$
You will then be able to put the numbers $B, C,$ and $E$ is size order. Together with the facts that $B, C, E < A$ and $A < D$, you will have the complete order.
