Joseph Bak and Donald Newman's complex analysis book (p.236) has a proof that the equation $e^z-z=0$ has infinitely many complex solutions:
I'm curious if there are any particularly elegant ways to see this, other than that given in the text.
Complex Analysis – Simple Ways to See That e^z – z = 0 Has Infinitely Many Solutions
complex-analysis
Best Answer
If you use the fairly deep result of Picard about essential singularities then you can prove this as follows: $f(z) = e^z-z$ has an essential singularity at infinity. Therefore $f$ attains all values infinitely many times with at most one exception (that is, at most a single value could be attained only finitely many times). This exception could still be $0$. However, $f$ also satisfies $f(z + 2\pi i) = f(z) - 2\pi i$. Now $f$ attains at least one value in $\{0, 2\pi i\}$ infinitely many times. In both cases it follows that $f$ must have infinitely many zeroes.