I have a calculation which ends up with the following form: $$a!(x!)^a=b!(y!)^b,$$
for $a,b,x,y$ non-negative integers. I would like to prove this is impossible in general unless $a=b$ and $x=y$ or some other trivial cases. W.l.o.g. $b>a$ and $x>1$. Then $x!^a=y!^b(b!/a!)=y!^b k$ for $k=b(b-1)\cdots(b-a+1)$. Since $b>a$ this already looks difficult to satisfy since we can write it as $$x^a(x-1)^a\cdots(x-y+1)^a = y!^{b-a}k$$ but I am not sure what step to take next.