Here is a sketch of a proof of the main ingredient:
$$\big(1+\frac{c_n}{n}\big)^n\xrightarrow{n\rightarrow\infty} e^c$$
for any $c_n\in\mathbb{C}$ with $c_n\xrightarrow{n\rightarrow\infty}c$.
Following Rick Durrett's approach to the classical CLT
Lemma I:
Let $\{z_j,w_j:1\leq j\leq n\}\subset\mathbb{C}$ such that
$\max_j\{|z_j|,|w_j|\}\leq \theta$. Then
\begin{align}
\Big|\prod^n_{j=1}z_j-\prod^n_{j=1}w_j\Big|\leq
\theta^{n-1}\sum^n_{j=1}|z_j-w_j|
\end{align}
Proof:
Let $a_0=\prod^n_{j=1}w_j$ and for $k\geq1$, $a_k=\Big(\prod^k_{j=1}z_j\Big)\Big(\prod^n_{j=k+1}w_j\Big)$. Then
\begin{align}
\Big|\prod^n_{j=1}z_j-\prod^n_{j=1}w_j\Big|&=\Big|\sum^n_{j=1}(a_j-a_{j-1})\Big|\leq\sum^n_{j=1}|a_j-a_{j-1}|\\
&=\sum^n_{j=1}|z_1\cdot\ldots\cdot z_{j-1}(z_j-w_j)w_{j+1}\cdot\ldots\cdot w_n|\leq \theta^{n-1}\sum^n_{j=1}|z_j-w_j|.
\end{align}
Lemma II:
Suppose $|z|\leq1$, then $|e^z-1-z|\leq|z|^2$.
Proof
Observe that $2^{n-1}\leq n!$ for $n\geq2$.
Since $e^z-1-z=\sum_{n\geq2}\frac{z^n}{n!}$ and $|z|\leq 1$,
it follows that
\begin{align}
|e^z-1-z|\leq \frac{|z|^2}{2}\sum_{n\geq2}2^{-(n-2)}=|z|^2
\end{align}
finally:
Proposition:
Suppose that $\{c_n\}\subset\mathbb{C}$ and that $c_n\rightarrow c$.
Then, $\big(1+\frac{c_n}{n}\big)^n\rightarrow e^c$.
Proof
Let $\gamma>|c|$ and $n_0$ large enough so that
$|c_n|<\gamma$ and $\gamma/n\leq 1$ whenever $n\geq n_0$.
If $z_j=1+\frac{c_n}{n}$ and $w_j=e^{c_n/n}$, $1\leq j\leq n$, then
have that
\begin{align}
\max_{1\leq j\leq n}\{|z_j|,|w_j|\}&\leq& e^{\gamma/n}\\
\big|e^{c_n/n}-1-\frac{c_n}{n}\big|&\leq& \frac{\gamma^2}{n^2}.
\end{align}
Therefore
\begin{align}
\big|(1+\frac{c_n}{n})^n-e^{c_n}\big|\leq e^{\gamma
(n-1)/n}n\frac{\gamma^2}{n^2}\rightarrow0.
\end{align}
The conclusion follows from the continuity of the exponential
function.
