Let $G$ be a finite group and $n_p:=|\text{Syl}_p(G)|$. Is it true that $n_p!\le |G|$ ?
I've shown that it's true, but I'm not so sure, can you check my proof?
Proof. Let $G$ act on $\text{Syl}_p(G)$ by conjugation, this action is well defined by the second Sylow theorem. Let $\pi:G\to \mathcal{S}(\text{Syl}_p(G))\cong\mathcal{S}_{\{1, \ldots, n_p\}}$ be this action. By the second Sylow theorem, $\forall i, j\in\{1, \ldots, n_p\}$ there exists $g\in G$ such that $\pi(g)(i)=j$, so $\pi$ is surjective (every permutation can be written as product of $2$-cycles) and $|G|\ge n_p!$.
The answer is no. With $G=A_4$ and $p=3$, $n_p=4$ but $|G|=12 < 4!$. You have not proved that $\pi$ is surjective, you have just proved that it is transitive, which means that $|G| \ge n_p$, which follows anyway from $n_p = |G/N_G(P)|$.
Your bracketed comment about every permutation can be written as a product of $2$-cycles does not make much sense. Also, you don't need Sylow's second theorem to show that $\pi$ is well defined.