I happened to read a statement related to my question in Ferguson's A Course in Large Sample Theory, where he stated:
$$\left(1 + \frac{\lambda_n(e^{it} - 1)}{n}\right)^n \to \exp[\lambda(e^{it} - 1)],$$ as $n \to \infty$, where $\lambda_n \to \lambda$ as $n \to \infty$.
So it looks like he used the property
$$\left(1 + \frac{z}{n}\right)^n \to e^z \tag{*}$$
as $n \to \infty$ tacitly, even when $z$ is a complex variable.
I understand that under the real setting, the proof of $e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$ relies on the ordering property of $\mathbb{R}$, but it is unlikely the same technique can be transferred to the complex domain.
Nevertheless, I guess this relation $(*)$ should be correct, but how do we show it rigorously (probably by using some complex analysis tools)?
I want to stress that this is an incomplete idea to go about this.
Assume that the function
$f(z)= \lim_{n \to \infty} \Big(1 + \frac{z}{n}\Big)^n - \exp(z)$
is analytic.
On the real line, it coincides with $0$ because $\lim_{n \to \infty} \Big(1 + \frac{x}{n}\Big)^n = \exp(x)$.
Since analytic functions can have at most countably many zeroes, it follows that $f(z) = 0$ for all $z \in \mathbb{C}$.