$z_0$ non-removable singularity of $f\Rightarrow z_0$ essential singularity of $\exp(f)$

We can also look at it from the other side.

If $z_0$ is a removable singularity of $e^f$, then $\lvert e^{f(z)}\rvert < K$ in some punctured neighbourhood of $z_0$. Since $\lvert e^w\rvert = e^{\operatorname{Re} w}$, that means $\operatorname{Re} f(z) < K'\; (= \log K)$ in a punctured neighbourhood $\dot{D}_\varepsilon(z_0)$ of $z_0$, and that implies that $z_0$ is a removable singularity of $f$. (Were it a pole, $f(\dot{D}_\varepsilon(z_0))$ would contain the complement of some disk $D_r(0)$; were it an essential singularity, each $f(\dot{D}_\varepsilon(z_0))$ would be dense in $\mathbb{C}$ by Casorati-Weierstraß; in both cases $\operatorname{Re} f(z)$ is unbounded on $\dot{D}_\varepsilon(z_0)$.)

If $z_0$ were a pole of $e^f$, it would be a removable singularity of $e^{-f}$, hence $z_0$ would be a removable singularity of $-f$ by the above, hence $z_0$ would be a removable singularity of $f$, and therefore a removable singularity of $e^f$ - contradiction.