Have you heard of central binomial coefficient?

If we use "\color\pink{?}" as relational operator, we get...

$(10! \: ?\: 5! * 5! *5!)$ $\Rightarrow {log_{10}(10!)}\:?\: {log_{10} (5!) +log_{10}(5!)+ log_{10}(5!)}$

$\Rightarrow{ log(10)+log(9)+\ldots +log(1)}\:?\: 3{ log (5)+log(4)+\ldots +log(1)} \:$ $\Rightarrow\left \lfloor{log(10)}\right\rfloor+\left\lfloor{log(9)}\right\rfloor+…+\left\lfloor{log(1)}\right\rfloor \:?\: 3{\left \lfloor {log(5)}\right\rfloor+\left\lfloor{log(4)}\right\rfloor+… +\left\lfloor{log(1)} \right \rfloor}$ $\Rightarrow 1+0+0……+0 \:?\: 3(0+0+……+0)$ $\Rightarrow 1>0$

So, it can be said that, \boxed{\color\red{10! \: > (5!)^3}}

We can write 5! as 5 * 4 * 3 * 2 * 1, and 10! as 10 * 9 * 8... * 1. Thus, 3 * 5! is 3 * 5 * 4 * 3... * 1. because 10 is greater than 5, 10! as all of the factors that 5! does. 10! can be written as 10 9 * 8... * 6 * 5!. Since each of the factors 10...6 is greater than 3, and since the factors 5...1 all cancel out, it is clear that 10! must be greater than 3 5!.