For example Chebyshev's theorem [aka Bertrand's postulate] that for $x > 1$ there is always a prime $x < p < 2x$. But you can also see that the product of all primes $\leqslant n$ is larger than $n$ for $n > 2$ (and much larger for all $n$ but the few smallest) without that.
What is the problem? If you know/accept that for all $n > 2$ there is a prime $p < n$ that doesn't divide $n$, do you see how that implies that $x^{x!} = (x!)^x$ can only be for $x \leqslant 2$?
No, you need that every prime that divides one side also divides the other. But for $x > 2$, there are always primes (at least one) dividing $(x!)^x$ but not $x^{x!}$.
Well, if $x\leqslant 2$ is ruled out by the problem statement, and we have proved that the two expressions are not equal for $x > 2$, then the answer is "none".
