Are there any simple ways to see that $e^z-z=0$ has infinitely many solutions?

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:

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I’m curious if there are any particularly elegant ways to see this, other than that given in the text.

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An elementary proof: Let $z = x +y i$ then $|e^z| = |z|$ precisely if $e^{2x} = x^2 + y^2$. If $x \geq 0$ then $e^{2x} – x^2 > 0$ so $y = (e^{2x} – x^2)^{1/2}$ is a positive solution of this equation. This means that for all $x \geq 0$ there is such a $y \geq 0$ such that $|e^z| = |z|$. The argument of $z$ is in $[0, \pi/2]$ since $x, y \geq 0$. The argument of $e^z$ is $y$. Since $y \to \infty$ when $x \to \infty$ there are infinitely many such $z$ for which both $|e^z|=|z|$ and $\arg(e^z) \equiv \arg (z) \pmod {2\pi}$. These $z$ are therefore roots of $e^z-z$.

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.

One way to see this is to realize that $f(z)=e^z$ has $\{z\in\mathbb{C}:0\leq\mathrm{Im}(z)<2\pi\}$ as a fundamental region and has period $2\pi i$. That fundamental region is mapped onto the plane (excluding $0$), as is every shift of the region by integer multiples of $2\pi i$. From there, it isn’t difficult to show that there must be infinitely many $z\in\mathbb{C}$ for which $f(z)=z$.

It remains only to show (as pointed out below by Harald) that in each such shift of the region there is at least one solution–that is, at least one zero of the function $g(z)=e^z-z$. Harald’s suggestion of applying the argument principle (see if needed) is a good one. Noting that the function is entire (so no poles), you really need only show that $\oint_C\frac{e^z}{e^z-z}dz$ is non-$0$ (where $C$ is the contour he suggests) for sufficiently large $M$.