I'm trying to find a formula for the product of factorials: $$\prod _{n=1}^{N}n!=\; ?$$ Now using a kind of "brute force", I believe that I can prove that $$\prod _{n=1}^{N}n!=\prod _{n=1}^{N}{n}^{N-n}$$ but I couldn't find a demonstration; I tried to use induction, but I only got $$\left( N+1 \right) \prod _{n=1}^{N}{\frac {{n}^{N}}{{n}^{n}}} $$

  • $\begingroup$ Considering the factorial operation is defined to concisely represent a product, I doubt there will be a nice expression. Can you explain what you've written at the end? I don't see how you've gone from an equation to just an expression. $\endgroup$ – preferred_anon Jul 7 '15 at 20:15
  • $\begingroup$ @Peterix Note that this product admit a closed form in terms of a special function. More precisely, we have $$\prod_{n=1}^{N}n!=\prod_{n=1}^{N}\Gamma\left(n+1\right)=G\left(N+2\right)$$ where $G$ is the Barnes $G$ function. $\endgroup$ – Marco Cantarini Jul 8 '15 at 12:07

Following @ZevChonoles, we have

$$\prod_{n=1}^{N}n!=\prod_{n=1}^{N}n^{N-(n-1)} \tag 1$$

We can prove this by induction. To that end, let's establish a base case. For $N=2$, we have



$$\prod_{n=1}^{2}n^{2-(n-1)} =(1^2)\,(2^1)=2$$

Now assume that $(1)$ is true for $N=K$. Then, examine

$$\begin{align} \prod_{n=1}^{K+1}n!&=(K+1)!\prod_{n=1}^{K}n!\\\\ &=(K+1)!\prod_{n=1}^{K}n^{K-(n-1)}\\\\ &=(K+1)!\prod_{n=1}^{K+1}n^{K-(n-1)}\\\\ &=(K+1)!\prod_{n=1}^{K+1}\frac{n^{(K+1)-(n-1)}}{n}\\\\ &=\prod_{n=1}^{K+1}n^{(K+1)-(n-1)} \end{align}$$

and we're done!

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  • $\begingroup$ I don't understand why $$(K+1)!\prod_{n=1}^{K}n^{K-(n-1)}=(K+1)!\prod_{n=1}^{K+1}n^{K-(n-1)}$$ $\endgroup$ – Peterix Jul 9 '15 at 12:16
  • $\begingroup$ @peterix We can multiply by $1$ and not change the answer. Put $K+1$ into the product and you get $1$. $\endgroup$ – Mark Viola Jul 9 '15 at 12:40
  • $\begingroup$ Sorry if I keep asking but I don't get why $$(K+1)!\prod_{n=1}^{K+1}\frac{n^{(K+1)-(n-1)}}{n}= \prod_{n=1}^{K+1}n^{(K+1)-(n-1)}$$ $\endgroup$ – Peterix Jul 9 '15 at 13:12
  • $\begingroup$ @Peterix $$\prod_{n=1}^{K+1}\frac{1}{n}=\frac{1}{(1)(2)(3)\cdots (K+1)}=\frac{1}{(K+1)!}$$ $\endgroup$ – Mark Viola Jul 9 '15 at 14:07
  • $\begingroup$ right, thanks a lot! $\endgroup$ – Peterix Jul 9 '15 at 14:08

Here is a simple demonstration, laid out pictorally: $$\begin{array}{c|ccccc} N!\strut_\strut & 1 & 2 & \cdots & (N-1) & N\\ (N-1)!\strut_\strut & 1 & 2 & \cdots & (N-1)\\ \vdots\strut_\strut & \vdots& &{\cdot}^{\Large\cdot^{\huge\cdot}} \\ 2!\strut_\strut & 1 & 2\\ 1!\strut_\strut & 1\\\hline \displaystyle\prod_{n=1}^Nn! & 1^N & 2^{N-1} & \cdots & (N-1)^2 & N \end{array}$$

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