Consider the functions $f_n(z)=\left(1+\frac{z}{n}\right)^n$ for each $n=1,2,3,...$ Use the fact that $f_n(x)$ $\rightarrow e^x $ for each $x \in \Bbb R $ to prove that $f_n(z) \rightarrow e^z$ normally in $\Bbb C$.
My work: Suppose $f_n(z)=\left(1+\frac{z}{n}\right)^n$ for each $n=1,2,3,...$ Since $1+x \le e^x$ for $x \ge 0$, it follows that $|f_n(z)| \le e^{|z|}$, for all $z$.
How can I show that $\{f_n \}$ is a normal family using Montel's theorem. Finally, my intuition is to prove $f_n(z) \rightarrow e^z$ normally in $\Bbb C$ using Arze-Ascoli theorem with the ideas of equicontinuity and pointwise bounded. We can prove it by contradiction. I don't know how to implement Arze-Ascoli theorem to show $f_n(z) \rightarrow e^z$ normally in $\Bbb C$ by contradiction. Your kind help will be appreciated. Thank you so much!
I'm not sure about one of your claims. I see
$$|f_n(z)|=|(1+z/n)^n|\le (1+|z|/n)^n \le e^{|z|},$$
the latter inequality being well known in calculus. I'm not sure how you obtained this.
Having this inequality shows that if $|z|\le R,$ then $|f_n(z)| \le e^R$ for all $n.$ Thus $f_n$ is uniformly bounded on compact subsets of $\mathbb C.$ This is looking good for Montel.
Suppose that $f_n(a)$ does not converge to $e^a$ for some $a\in \mathbb C$ Then for some $\epsilon>0,$ there is a subsequence $n_k$ such that
$$\tag 1 |f_{n_k}(a) - e^a| \ge \epsilon, \, k=1,2,\dots$$
Let $R=|a|+1.$ Because $f_{n_k}$ is uniformly bounded on $D(0,R),$ Montel says there is a further subsequence $f_{n_{k_j}}$ that converges uniformly on $D(0,|a|+1/2).$
Let $g$ be the uniform limit of $f_{n_{k_j}}$ on this disc. Then $g$ is holomorphic there, and equals $e^x$ on the real axis intersected with this disc by the given hypothesis. By the identity principle, $g(z)= e^z$ in $D(0,|a|+1/2).$ This implies $f_{n_{k_j}}(a) \to e^a,$ contradicting $(1)$ and proving our result.