Question: Are there any useful forms for the expression $1!\cdot 2!\cdot 3!\cdot ...\cdot n!$? I'm trying to solve a problem that involves this expression and thought it might help to find a more "workable" form for it, but I didn't get very far.

Thanks

Answer: 
A posibble form is: $$\prod\_{i=1}^ni!$$

In other way is $$1^n\cdot2^{n-1}\cdot3^{n-2}\cdot\cdots\cdot(n-1)^2\cdot n^1$$
Let's get the asymptotic behavior of $\prod\_{k=1}^n k! = \prod\_{k=1}^n k^{n-k+1}$.

The standard way to estimate such things is by taking the logarithm. But in this case, we can get it directly from Stirling's approximation (which itself can be proved by taking logs).

[Stirling's approximation](http://en.wikipedia.org/wiki/Stirling's_approximation) states $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$.

Thus $\prod\_{k=1}^n k! \sim (2\pi)^n \sqrt{n!} \frac{\prod\_{k=1}^n k^k}{e^{n(n+1)/2}}$

So now we just need to estimate $\prod\_{k=1}^n k^k$. We could go do this, but even better is to use:

$\prod\_{k=1}^n k! \prod\_{k=1}^n k^k = \prod\_{k=1}^n k^{n-k+1} \prod\_{k=1}^n k^k = \prod\_{k=1}^n k^{n+1} = (n!)^{n+1}$

Thus $(\prod\_{k=1}^n k!)^2 \sim (2\pi)^n \sqrt{n!} \frac{(n!)^{n+1}}{e^{n(n+1)/2}}$

Then we get, applying Stirling's approximation to $\sqrt{n!}$ and $(n!)^{n+1}$:

$(\prod\_{k=1}^n k!)^2 \sim (2\pi)^{2n+3/2}n^{3/4 + n/2}e^{-n(n+1)/2-n/2-n(n+1)} n^{n(n+1)}$

Thus $\prod\_{k=1}^n k! \sim (2\pi)^{n+3/4} e^{-3n^2/4 - n} n^{n^2/2 + 3n/4 + 3/8}$