I was trying to find a way to find solutions to the factorial equation $(xy)!=x!y!$, where $x,y$ are nonnegative integers. I began with the case when $x=y$. This leads to
\begin{equation} (x^2)!=(x!)^2 \end{equation} which is equivalent to \begin{equation} x^2(x^2-1)(x^2-2)!=x^2(x-1)^2(x-2)!^2\Leftrightarrow x^2(x-1)[(x+1)(x^2-2)!-(x-1)(x-2)!^2]=0\Leftrightarrow x=0,1. \end{equation} This case was quite easy to solve. The problem comes when considering the case when $x<y$. I suspect that the solutions for this case will be given when $x=1$ and $y>1$. Any suggestions or hints that you can give me in order to solve it for this particular case will be really appreciated. Thank you in advanced.
Hint: $\frac {(xy)!} {x!} >(x+1)(x+2)...(x+y)>(1)(2)...(y)=y!$ if $x,y \geq 1$ and $xy \geq x+y$. Investigate when $xy \geq x+y$ holds. [This is quite trivial!]