I might have solved it but I don't know if the methods I used are valid. Please check my work.

$$z=4e^z$$
I apply $z^{-1}$ to both sides to get
$$\frac{1}{z}=\frac{1}{4}e^{-z}$$

now that's the step I'm worried about. $z$ can't be $0$ for me to do that, and it's easy enough to check that $z$ indeed isn't one of the solutions, but is that enough justification?

In any case, I continue with multiplying both sides with $z$ to get

$$1=\frac{1}{4}ze^{-z}$$

Now applying Rouche's theorem for $f(z)=1$ and $g(z)=-\frac 1 4 ze^{-z}$ I find that there are no solutions inside the unit circle.

Am I allowed to take those two steps? If yes, how can I be confident that I didn't perhaps introduce new solutions, or wipe out existing ones? If not, why not?