If $z_n \to z$ then $(1+z_n/n)^n \to e^z$

We are dealing with $z \in \mathbb{C}$. I know that $$ \left(1+ \frac{z}{n} \right)^n \to e^{z} $$ as $n \to \infty$. So intuitively if $z_n \to z$ then we should have $$ \left(1+ \frac{z_n}{n} \right)^n \to e^{z}. $$ If $z_n \in \mathbb{R}$ I would be happy writing $$ \exp \left(n \log\left(1+ \frac{z_n}{n} \right) \right) = \exp \left(z_n \frac{\log\left(1+ \frac{z_n}{n} \right)-\log(1+0)}{\frac{z_n}{n}-0} \right) \to \exp(z \cdot 1) $$ where here we are using the definition of derivative. But if $z \in \mathbb{C}$ the logarithm is multivalued and I'm not sure that the same calculation is allowed. Is there a different way to show this for complex $z$?

You do not need the logarithm function at all.

We begin with the bound, valid for complex $z$ with $|z|\leq 1$: $$|(1+z)-\exp(z)|\leq \left|{z^2\over2!}+{z^3\over 3!}+\cdots\right|\leq {|z|^2\over 2!}+{|z|^3\over 3!}+\cdots\leq |z|^2.$$ Similarly, we also have $|1+z|\leq \exp(|z|)$ and $|\exp(z)|\leq \exp(|z|)$ for all $z$.

Now suppose that $c_n\to c$ in the complex plane. Consider the telescoping sum $$w_1\cdots w_n-z_1\cdots z_n=\sum_{j=1}^n w_1\cdots w_{j-1}(w_j-z_j)z_{j+1}\cdots z_n,$$ and plug in $w_j=(1+c_n/n)$ and $z_j=\exp(c_n/n)$ to obtain $$\left(1+{c_n\over n}\right)^n-\exp(c_n)= \sum_{j=1}^n \left(1+{c_n\over n}\right)^{j-1}\left[\left(1+{c_n\over n}\right)-\exp(c_n/n)\right]\exp(c_n/n)^{n-j}.$$ For $n$ so large that $|c_n/n|\leq 1$, the bounds above give $$\left|\left(1+{c_n\over n}\right)^n-\exp(c_n)\right|\leq n \exp(|c_n|)\, {|c_n|^2\over n^2}\to 0\mbox{ as }n\to\infty.$$

This shows that $\left(1+{c_n\over n}\right)^n\to\exp(c)$ as $n\to\infty.$

For your argument to be valid, specify a branch of $\log$; You can use the principal branch, $\text{Log}(z)$, but any branch that is analytic at $z=1$ should do:

$$\lim_{n\to 0}\left( 1 + \frac{z_n}{n}\right)^n = \lim_{n \to 0} \exp\left( n \text{ Log}\left( 1 + \frac{z_n}{n}\right)\right) = \cdots$$

You can use the following standard lemma:

Lemma: If $a_{n} $ is a sequence of complex numbers such that $n(a_{n} - 1)\to 0$ then $a_{n} ^{n} \to 1$.

Now let's put $$a_{n} =\dfrac{1+\dfrac{z_{n}}{n}}{1+\dfrac{z}{n}}$$ so that $$n(a_{n} - 1)=\frac{z_{n}-z}{1+z/n}\to 0$$ and then $a_{n} ^{n} \to 1$ and this means that $(1+(z_{n}/n))^{n}$ tends to the same limit as that of $(1+(z/n))^{n}$ and we are done.

The lemma mentioned at the beginning is proved easily by writing $a_{n} =1+b_{n}$ so that $nb_{n} \to 0$ and we have via binomial theorem $$|a_{n} ^{n} - 1|=|(1+b_{n})^{n}-1|=\left|nb_{n}+\frac{n(n-1)}{2!}b_{n}^{2}+\dots\right|$$ and clearly the RHS is bounded above by $$|nb_{n} |+|nb_{n} |^{2}+\cdots = \frac{|nb_{n} |} {1-|nb_{n}|}$$ Since $nb_{n} \to 0$ we are done.