$\sin z=z$ has infinitely many solutions in $\CC$

Let \(f(z) = \sin(z)-z\) and apply Big Picard: \(f\) takes on every value in \(\CC\), except possibly for some single \(z_0\), infinitely many times. So there is at most one \(z_0\) such that there are finitely many solutions to \(\sin(z) - z = z_0\). If \(z_0\neq 0\), we’re done, so suppose such an exception \(z_0\) does exist whose fiber has finitely many points. Since \(z_0+2\pi \neq z_0\), there must then be infinitely many solutions to \(f(z) = z_0 + 2\pi\). But then \[ f(z+2\pi) = \sin(z+2\pi) - z - 2\pi = \sin(z) - z - 2\pi = (z_0 + 2\pi)-2\pi = z_0 ,\] so we can produce infinitely many points in the fiber over \(w_0\), contradicting finiteness. So no exception \(z_0\) exists, and in particular, \(z_0=0\) is not an exception so \(f(z) = 0\) has infinitely many solutions.