Prove Picard's theorem for $f(\zeta)=e^{1/\zeta}$.
Picard's theorem: If $f$ has essential singularity at $c$, then $f(D_r(c)=\mathbb{C}$ or $=\mathbb{C}\backslash \{1 point\}$
Proof: The Laurent series for $f(\zeta)=e^{1/\zeta}$ is $$f(\zeta) =\sum\limits_{n=0}^\infty \dfrac{(1/\zeta)^n}{n!} =\sum\limits_{n=0}^\infty \dfrac{1}{n!}\zeta^{-n} $$ It is clear that $f$ has an essential singularity at $0$ since $a_n\neq 0$ for infinitely many $n>0 \iff -n<0$. Let $D_r(0)$ be a disk centered at the singularity with radius $r>0$. Note that $f(\zeta)\neq 0$ for any $\zeta \in D_r(0)$ no matter how large we make $r$. This implies that $f(D_r(0))=\mathbb{C}\backslash\{0\}$.
This seems so easy. Did I do the proof correctly?
We cannot "prove Picard's theorem for the function $f(\zeta):=e^{1/\zeta}\>$", but we can prove that for any $r>0$ the restrictionn $f\restriction D_r$ assumes all nonzero complex numbers as values.
Note that the function $z\mapsto e^z$ assumes all nonzero complex numbers as values in any horizontal strip of width $>2\pi$. Since the preliminary map $\zeta\mapsto z:=1/\zeta$ maps $D_r$ onto the exterior of $D_{1/r}$, and this exterior contains an infinity of full such strips, the claim follows.