Let $x,y,n$ be postive integers such that $x\ge 2y,y>n,n>3$
I conjectured that $$\color{red}{x!-y!\ge x^n}$$
Now, I claim that $$\color{red}{x!-y!=y![x(x-1)(x-2)\cdots(y+1)-1]\ge (n+1)![x(x-1)(x-2)\cdots (y+1)-1]} \tag{1}$$
and I can prove $$\color{blue}{(k+2)(x-k)\ge x,k<n,x\ge 2n}$$ So we have $$\color{blue}{(n+1)!x(x-1)(x-2)\cdots (x-n+1)\ge x^n}\tag{2}$$which I think, gives little success.
$\color{blue}{Reason}$ since $$\color{purple}{x(x-1)(x-2)\cdots(y+1)<x(x-1)(x-2)\cdots (x-n+1)}$$ Combing $(1),(2)$ I can't prove it. What is the solution to this problem?
The given inequality is $x!-y!-x^n \geq 0$
Observe that $x! \geq 2(x-1)!$
So if we can prove, $[(x-1)!-y!]+[(x-1)!-x^n] \geq 0$, we are done.
$(1)$ $(x-1)!-y! \geq (2y-1)! -y! \geq 0$ (as $x \geq 2y$)
$(2)$ $(x-1)! -x^n \geq (x-1)! -x^{(x-2)/2}$
[as $x \geq 2y, y >n \Rightarrow x \geq 2n+2 \Rightarrow n \leq \frac{x-2}{2}$]
So we are left to prove $(x-1)! > x^{(x-2)/2} \Leftrightarrow (x!)^2 > x^x$, which is true.