= Solution
Let $F=e^g$. It is <holomorphic> and nowhere zero on a punctured neighbourhood of $a$, and the <chain rule> gives
$$
\frac{F'}F=g'.
$$
If $F$ had a <pole> of order $n\geq1$, part (c) would imply $\operatorname{Res}_a(g')=-n$. But the derivative of a <holomorphic function> on a punctured disc has <residue> zero. Indeed, write its <Laurent series>
$$
g(z)=\sum_{k\in\mathbb Z}c_k(z-a)^k,\qquad g'(z)=\sum_{k\in\mathbb Z}k c_k(z-a)^{k-1}.
$$
Termwise differentiation is valid on every smaller closed annulus. The only possible $(z-a)^{-1}$ term comes from $k=0$ and has coefficient zero. Thus
$$
\boxed{\operatorname{Res}_a(g')=0\ne-n},
$$
a contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves <an exponential has no isolated pole>, including when $g$ has an <essential singularity>; no choice of a <holomorphic logarithm> on the entire punctured disc was assumed.
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