How to prove $\cos(n!) \neq 1$ without using $\pi$ is irrational Prove
$[\forall n \in \mathbb{N}, \cos(n!) \neq 1]$,
without using $\pi$ is irrational.
Using $\pi \in (\mathbb R-\mathbb Q)$, I can prove it...

Thanks everyone!

This is not a full answer, just a comment written to give someone more skilled an idea of possible proof.
If we have $\cos (n_0!)=1$ for some $n_0 \in \mathbb N$ then we have $(\cos (n_0!))^2=1$. This implies $\sin(n_0!)=0$ and this implies $$\sum_{n=0}^{+ \infty}\frac{(-1)^n \cdot (n_0!)^{2n+1}}{(2n+1)!}=0$$ and this probably could be shown to be impossible.
