I know how to prove $(1+x/n)^n \to e^x$ uniformly if $x \in [a,b] \subset \mathbb{R}$ by finding a uniform bound $e^x - (1 +x/n)^n < K/n$ for a bounded interval.
My question is how to prove the $(1 + z/n)^n \to e^z$ on any compact set $S \subset \mathbb{C}$. I am not sure how to extend from the real case in the proof.
Let Log denote the principal branch of logarithm. Then $\frac {Log (1+z)} z \to 1$ as $z \to 0$. Let $\epsilon >0$. There exists $\delta >0$ such that $|\frac {Log (1+z)} z -1| < \epsilon$ if $|z| <\delta$. Let K be a compact set in $\mathbb C$. Then $|n \frac {Log (1+z/n)} z -1| <\epsilon$ for all $z \in K$ provided n is sufficiently large. [ Because $|\frac z n|<\delta$ if $z \in K$ and n is sufficiently large]. Hence $ |{Log \{(1+z/n)^{n}\}} -z| <\epsilon |z|$ By continuity of the exponential function at 0 this proves that$(1+z/n)^{n} e^{-z} \to 1$ uniformly on K. Finally, $|(1+z/n)^{n}-e^{z}| \leq |(1+z/n)^{n} e^{-z}-1| e^{|z|} \to 0$ uniformly on K.