Solutions of $((a-1)!)^x=0$

Question

I have an equation that I couldn't solve. I will be glad if you help me to do it.

Are there solutions in: $((a-1)!)^x=0,$ with $a\in\mathbb N?$

Because with my knowledge in math, I found no solutions.

Answer 1

$$n!\geq 1 \ \forall \ n \in \mathbb N$$

Substituting $n$ with $n-1$ does not change this fact, so your equation simplifies to $g^x = 0$, with $g \in \mathbb N$. There does not exist an $x$ value in $ \mathbb R$ (and I am nearly confident there does not exist any value in $\mathbb C$ either, but perhaps I am mistaken) that satisfies that equation, so you are correct in saying there are no solutions given your guidelines.

Edit: I am now fully confident (and backed up by Joe's comment) that there does not exist an $x \in \mathbb C$ that satisfies the equation.