Showing that $e^z=z$ has infinitely many solutions [duplicate]

One of our review problems for my complex analysis final is the following:

Show that $e^z=z$ has infinitely many solutions in $\mathbb{C}$.

I've seen a few solutions online, but none of them have used material that was covered in our introductory complex analysis course. Would anyone be able to provide a simple solution?

I'm trying to think of ways to invoke bounding in some way but I can't seem to come up with anything.

From page 5 of this book :

By Picard’s big theorem and a previous example in the book, we know $e^z$ has an essential singularity at infinity, hence so does $f(z) = e^z − z$. This implies that as $|z| → ∞$, we know $f(z)$ takes on all values infinitely many times with the possible exception of one point. This point could still be zero; however $f(z + 2πi) = f(z) − 2πi$. Therefore, we know $f(z)$ takes on at least one of ${0, 2πi}$ infinitely many times, hence has infinitely many zeros.

Let $k \in \mathbb{N}$, $k\geq 1$ and let $\log_k$ be the branch of $\log$ that maps the upper half plane onto the strip $H_k = \{z \mid 2k\pi < \operatorname{Im}(z) < (2k+1)\pi\}$. Then $\log_k$ is a contraction mapping of $\overline{H}_k$ (since $|\log_k'| \leq (2k \pi)^{-1}<1$ on that closed strip). Therefore $\log_k$ has a single fixed point $z_k \in \overline{H}_k$. Clearly $\exp(z_k)=z_k$.