The function e^(1/z) omits zero and takes every nonzero value infinitely often near the origin

Since the exponential never vanishes, $f(z)$ never equals $0$ on the punctured disc.

Fix $w\mathrm{\ne }0$. By [L1] and [L2], choose $\lambda \in \mathbb{C}$ with $e^{\lambda }=w$; then every number $\lambda +2\pi ik$ is another logarithm of $w$. Since $\mathrm{\mid }\lambda +2\pi ik\mathrm{\mid }\to \mathrm{\infty }$ as $\mathrm{\mid }k\mathrm{\mid }\to \mathrm{\infty }$, for every sufficiently large $\mathrm{\mid }k\mathrm{\mid }$ it exceeds $1$. For those infinitely many indices set $z_k:=1\mathrm{/}(\lambda +2\pi ik)$; then $0<\mathrm{\mid }{z}_k\mathrm{\mid }<1$, $f(z_k)=w$, and $z_k\to 0$. Thus every nonzero value occurs infinitely often within the punctured-disc domain near $0$.

Along $z=1\mathrm{/}n$, $f(z)=e^n\to \mathrm{\infty }$, whereas along $z=-1\mathrm{/}n$, $f(z)=e^{-n}\to 0$. Both sequences eventually lie in the punctured disc, so $0$ is neither removable nor a pole, and hence is an essential singularity.

Therefore Great Picard is sharp: an essential singularity can omit one finite value, namely $0$.