Is $n!\alpha \bmod 1$ dense in $[0,1]$?

Question

We know that positive integer times a irrational number modulo $1$ generate a dense set in $[0,1]$. According the answer of this post:Multiples of an irrational number forming a dense subset. I see no reason why the proof cannot be extended to $n!\alpha$ for $\alpha$ be an irrational number. We can just replace $i$ and $j$ with $i!$ and $j!$ and the argument still holds. Is that true?

Answer 1

I presume you mean $n!\alpha$ (not $n!/\alpha$).

Try $\alpha=e$. (Yes, that $e$.) Then $$n!e=\text{integer}+\frac1{n+1}+\frac1{(n+1)(n+2)}+\cdots$$ so modulo $1$, $n!e$ is between $1/(n+1)$ and $1/n$, so the $n!e$ are certainly not dense modulo $1$ in $[0,1]$.