Behavior of $e^{ f(z)}$ if f has removable singularity at a

This particular problem was asked in my assignment which could not be discussed due to pendamic.

Question : If f is holomorphic on $$\Omega$$/{a} prove that $$e^{f(z)}$$ cannot have a pole at a .

What I tried :If f has a pole of order k at a and let $$|e^{f(z)}|\to \infty$$ as z$$\to$$ a. then I took $$g(z)=\frac{f(z)} {(z-a)^k}$$. Then $$\frac{g(z)}{(z-a)^k} \to \infty$$ as z$$\to a$$ .( But I don't know how to proceed in this case from here).

Let f have an isolated singularity. I am confused in this part on which property should be used to prove that pole does not exists .

If f has an essential singularity at a then also , I am confused on which result to use . Can it be proved that $$e^{f(z)}$$ will also have an isolated singularity at z$$\to$$ a ? If yes kindly give some hints .

If singularity is pole then it has an answer here:pole becomes essential singularity when lifting by exponential

If singularity is essential then it has an answer here:If $z=a$ is not the removable singularity of $f$, show that $e^{f(z)}$ has essential singularity at $z=a$.

But I am not able to find an answer when it's removable singularity .

Can anyone please answer it ? I shall be really thankful as I am not good in dealing with singularities.

Thanks

If $$e^f$$ has a pole at $$a$$, then $$|e^f|$$ goes to infinity at $$a$$. Thus the real part of $$f$$ must go to infinity at $$a$$. This shows that, at $$a$$, $$f$$ cannot have a removable singularity, nor an essential singularity (by, eg, Casorati-Weierstrass – in the neighborhood of an essential singularity, a dense subset of $$\mathbb{C}$$ is reached infinitely many times). Thus $$f$$ has a pole.
Write $$f(z)=\frac{g(z)}{(z-a)^k}$$ with $$g$$ holomorphic at $$a$$ and $$g(a) \neq 0$$. Let $$\tau$$ be such that $$R=g(a)e^{-i\tau} > 0$$. Then, with $$z=a+\epsilon e^{i(\pi+\tau)/k}$$, $$f(z)$$ is $$-\epsilon^{-k}(R+o(1))$$ so its real part goes to $$-\infty$$, a contradiction.