I tried and I got there doesn't exist such a $n_0$
However, I dun think I have a formal proof for this.
My approach is, First assume there is such a $n_o$ exist and start my calculation with induction
$(n+1)!\\ =(n+1)n!\\ \gt(n+1)2n^3 (Induction Premise)\\ = 2n^4 + 2n^3\\ \gt 2n^3+6n^2+6n+1 , n \geq 3 \\ = 2(n+1)^3 $
So I get $n_0$ = 3, But when I subsitute back to the original equation, it doesn't work, so, what can i conclude? It doesn't exist such a $n_0$?
You need a base case. You're right that $n_0=3$ doesn't work, but that does not mean that there is no $n_0$ that works. What you've shown is that for your inductive step to work, you will need $n$ to be at least $3$, but there is still the requirement that you get the inequality to hold once in the first place.
So, one way to proceed is to find an $n_0>3$ for which the inequality holds, then prove by induction that the inequality holds for all $n\geq n_0$.
In fact, you can make your induction step a little easier by taking $n_0$ (very) slightly larger than necessary. If $n\geq 7$, then $2(n+1)^3<2(2n)^3=8(2n^3)\leq(n+1)2n^3$.