Product of the first $N$ factorials

Question

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 $$\left( N+1 \right) \prod _{n=1}^{N}{\frac {{n}^{N}}{{n}^{n}}} $$

Answer 1

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

Answer 2

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

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

and

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

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!