TITLE: Product of the first $N$ factorials
QUESTION [6 upvotes]: 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  $$\left( N+1 \right) \prod _{n=1}^{N}{\frac {{n}^{N}}{{n}^{n}}}
$$

REPLY [5 votes]: 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!