Let $f(z)=e^{- \frac{1}{z}}$. I would like to verify that $\forall R > 0, \ f(B(0,R)-\{0\})= \mathbb{C} - \{0\}$ without the use of the Casorati-Weierstrass theorem. $B(0,R)=\{x+iy \in \mathbb{C} \ | \ \sqrt{x^2+y^2} < R\}$ would be the open ball or radius $R$ and center $0$.
At first, I thought I may try to prove it by contradiction, therefore I'd suppose that $\exists R > 0$ such that $F(B(0,R)-\{0\}) \not\ni z_0$, where $z_0 \in \mathbb{C}-\{0\}$. At this point, I thought I could maybe study the singularities of the function $g(z)= \dfrac{1}{f(z)-z_0}$ and possibly deduce some absurdity about the singularities of $f(z)$. Unfortunately, I am unable to conclude.
Would you reason the way I did or is there another method you'd advise me to use?
It also came to my mind that I could try proving it directly by studying the range of $f(z)$, but it looked to me like I had to get too much in detail about the radiuses of the balls, so I gave up.
Observe that, if $g(z)=-1/z$, $R>0$, $U_R=B(0,R)\!\smallsetminus\!\{0\}$ and $W_R=\{z\in\mathbb{C}: \mathrm{Im}\,z>R\}$, then $$ W_{1/R}\subset g[B_R]. $$ Next observe that $$ \exp[W_r]=\mathbb C\!\smallsetminus\!\{0\},\quad \text{for all $r>0$}. $$