Exponential Sequence is Uniformly Convergent on Compact Sets

Theorem
Let $\mathcal E = \left \langle{E_n}\right \rangle$ be the sequence of functions $E_n: \C \to \C$ defined by $E_n(z) = \left({1 + \dfrac{z}{n}}\right)^n$.

Let $K$ be a compact subset of $\C$.

Then $\mathcal E$ is uniformly convergent on $K$.

$\mathcal E$ is Uniformly Bounded
First, from Equivalence of Definitions of Complex Exponential   we see that $\mathcal E$ is  pointwise convergent to $\exp$.

Now, since $K$ is compact,  it is bounded. Let $M$ be this bound and let $z \in U$, where $U$ is the open disk of radius $M + 1$.

Pick some arbitrary $n \in \N$.

Observe that:
 * $\left|{1 + \dfrac{z}{n}}\right| \leq 1 + \dfrac{\left|{z}\right|}{n}$

by the Triangle Inequality, so:
 * $\left|{1 + \dfrac{z}{n}}\right| ^n \leq \left({1 + \dfrac{\left|{z}\right|}{n}}\right) ^n$

Also:
 * $\left| {z} \right| \leq M + 1 \implies \left({1 + \dfrac{\left|{z}\right|}{n}}\right) ^n \leq \left({1 + \dfrac{\left|{M+1}\right|}{n}}\right) ^n $

Thus it is sufficient to show that $\left({1 + \dfrac{\left|{M+1}\right|}{n}}\right) ^n$ is bounded.

Also note that $E_n(z)$ is holomorphic on $U$.

From Exponential Sequence is Eventually Increasing, we have that:
 * $\exists N \in \N : n \geq N \implies \left({1 + \dfrac{\left|{M+1}\right|}{n}}\right) ^n < \left({1 + \dfrac{\left|{M+1}\right|}{n+1}}\right) ^{n+1}$

So, for $n \geq N$:
 * $\displaystyle \left|{1 + \dfrac{M+1}{n}}\right| ^n \leq \lim_{n \to \infty}\left({1 + \dfrac{\left|{M+1}\right|}{n}}\right) ^n = \exp \left({M+1}\right)$

Now, for $n \in \left\{{1, 2, \ldots, N-1}\right\}$


 * $\exists K_n \in \R: \left({1 + \dfrac{\left|{M+1}\right|}{n}}\right) ^n \leq K_n$

by Continuous Function on Compact Space is Bounded.

In summary, $\forall n \in \N: E_n$ is bounded by $\max \left\{{K_0, K_1, \ldots, K_{N-1}, \exp(M+1)}\right\}$ on $U$.

That is, $\mathcal E$ is uniformly bounded on $U$.

$\mathcal E$ is Uniformly Convergent on K
Since $\mathcal E$ is uniformly bounded on $U$,  $\mathcal E$ is locally bounded  on $U$.

By Montel's Theorem, $\mathcal E$ is a  normal family, so by  Vitali's Convergence Theorem, $\mathcal E$ converges uniformly on compact subsets of $U$. In particular, $\mathcal E$ converges uniformly on $K$.