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

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:

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

• Your argument is fine, because $1+z_n/n\to 1$ necessarily, and so you only need $\log$ to be well behaved locally. Picking a consistent branch works fine.
• On the other hand, you could 'squeeze' the result, by noting that the modulus difference from $e^z$ decays as $z_n\to z$. This argument works by continuity. (Edit: Fixed from only working for the real case, d'uh!)