Formula for $1! \times 2! \times \cdots \times n!$?

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 1

In the May 2013 Fibonacci Quarterly (Vol. 51, Num. 2) pages 163-173, Michael Hirschhorn proved this result:

Let $P(n) =\prod\limits_{k=0}^n \binom{n}{k} $. Then (this is going to be a pain to enter), as $n \to \infty$,

$$P(n) \sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} \exp\left(-\sum\limits_{p \ge 1}\dfrac{B_{p+1}+B_{p+2}}{p(p+1)}\dfrac1{n^p}\right) $$ where $$C =\lim\limits_{n \to \infty} \dfrac1{n^{1/12}}\prod\limits_{k=1}^n\left( \dfrac{k!}{\sqrt{2\pi k}(k/e)^k} \right) \approx 1.046335066770503 $$ and the $\{B_p\}$ are the Bernoulli numbers, defined by $$ \sum\limits_{p \ge 0} B_p \dfrac{x^p}{p!} = \dfrac{x}{e^x-1} .$$

I will now try to use this to get an estimate for $f(n) =\prod\limits_{k=0}^{n} n! $.

Since $P(n) =\dfrac{(n!)^{n+1}}{\prod\limits_{k=0}^{n} (n!)^2} $, $f(n) =\prod\limits_{k=0}^{n} n! =\sqrt{\dfrac{(n!)^{n+1}}{P(n)}} $.

I will now try to get an estimate for $f(n)$ from the first term ($p=1$) in the asymptotic formula for $P(n)$..

Since $B_2=\frac16$ and $B_3=0$, $B_2+B_3=\frac16$, so, setting $p=1$ in the formula for $P(n)$,

$\begin{array}\\ P(n) &\sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} e^{-1/(12n)}\\ &\sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} \\ \end{array} $

since that $e^{1/(12n)}$ term goes to $1$.

Therefore $\dfrac1{\sqrt{P(n)}} \sim C^{1/2}\dfrac{n^{(3n+2))/12}(2\pi)^{(2n+1)/8}}{e^{n(n+1)/4}} $.

Since $n! \sim \sqrt{2\pi n}\left( \dfrac{n}{e}\right)^n $, $(n!)^{(n+1)/2} \sim (2\pi)^{(n+1)/4} n^{(n+1)/4}\dfrac{n^{n(n+1)/2}}{e^{n(n+1)/2}} = \dfrac{(2\pi)^{(n+1)/4}n^{n(n+1)/2+(n+1)/4}}{e^{n(n+1)/2}} $.

Therefore,

$\begin{array}\\ f(n) &=\prod\limits_{k=0}^{n} n!\\ &=\sqrt{\dfrac{(n!)^{n+1}}{P(n)}}\\ &\sim \dfrac{(2\pi)^{(n+1)/4}n^{n(n+1)/2+(n+1)/4}}{e^{n(n+1)/2}} C^{1/2}\dfrac{n^{(3n+2))/12}(2\pi)^{(2n+1)/8}}{e^{n(n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(n+1)/4+(2n+1)/8}n^{n(n+1)/2+(n+1)/4+(3n+2)/12}}{e^{n(n+1)/2+n(n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n(n+1)/2+(6n+5)/12}}{e^{n(2n+2+n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+(12n+5)/12}}{e^{n(3n+4)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+(12n+5)/12}}{e^{3n^2/4+n}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+n+5/12}}{e^{3n^2/4+n}}\\ &= C^{1/2} (2\pi)^{3/8}n^{5/12}(2\pi)^{n/2}(n/e)^n \left(\dfrac{n}{e^{3/2}}\right)^{n^2/2}\\ \end{array} $

(Yep, that definitely was a pain to enter.)

As usual, since the math was done as it was entered, prob(no error) < .5.

Answer 2

A posibble form is: $$\prod_{i=1}^ni!$$

In other way is $$1^n\cdot2^{n-1}\cdot3^{n-2}\cdot\cdots\cdot(n-1)^2\cdot n^1$$

Answer 3

It is sometimes referred to as the superfactorial of n (although there is at least one other function that's also called superfactorial), and its values are given by OEIS sequence A000178. I don't believe there are any shortcuts to computing it, other than noting that $sf(n) = \prod_{1 \le i \lt j \le n}\left( j - i \right)$

Answer 4

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