Exponential Sequence is Uniformly Convergent on Compact Sets

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

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

Then $E_n$ is uniformly convergent on $K$.

Proof
First, from Equivalence of Definitions of Complex Exponential   we see that $E_n$ 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:
 * $\left|{1 + \frac{z}{n}}\right| \leq 1 + \frac{\left|{z}\right|}{n}$

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

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

From Exponential Sequence is Increasing, we have that:
 * $\left({1 + \frac{\left|{z}\right|}{n}}\right) ^n < \left({1 + \frac{\left|{z}\right|}{n+1}}\right) ^{n+1}$,

so:
 * $\displaystyle \left|{1 + \frac{z}{n}}\right| ^n \leq \lim_{n \to \infty}\left({1 + \frac{\left|{z}\right|}{n}}\right) ^n = \exp \left|{z}\right|$.

Since Exponential is Strictly Increasing, this means that $\exp \left|{z}\right| \leq \exp \left({M+1}\right)$.

In summary, $\forall n \in \N: E_n$ is bounded on $U$. That is, $\mathcal E$ is uniformly bounded and thus   $\mathcal E$ 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 $K$.