Zeros of $\sinh z = z$ in $\mathbb C$

Question

I must show that $f (z)= \sinh z - z$ has infinite zeros in $\mathbb C $, and that all zeros (except $0 $) are simple. Since the order of growth of $f $ is $1$, if it had finitely many zeros we'd have $f (z)=p (z) e^{az+b}$. Since if $z $ tends to $\infty $, $f\sim e^z$, and if $z $ tends to $-\infty $, $f \sim e^{-z}$. This would mean that $p(z) $ is constant, and that Re$ (a)$ is equal to both $1$ and $-1$, clearly absurd. However, I have no idea on how to prove that the nonzero zeros of $f $ are simple. Plus, I don't understand well why the proposition written in bold holds. Thanks for any clarify.