=== $\mathcal E$ is Uniformly Bounded on an open space containing $K$ === First, from Equivalence of Definitions of Complex Exponential Function  we see that $\mathcal E$ is pointwise convergent to $\exp$. {{refactor|extract this result from that page}} From Combination Theorem for Continuous Functions, $E_n$ is continuous for each $n \in \N$. From Compact Subspace of Metric Space is Bounded, $K$ is bounded by some real number $M \in \R$. Let: : $M' = M + 1$ : $U = \left \{{z \in \C: \left \vert{z}\right \vert < M'}\right \}$ : $z \in U$ : $n \in \N$ From the Triangle Inequality for Complex Numbers: :$\left \vert{1 + \dfrac z n}\right \vert \le 1 + \dfrac {\left \vert{z}\right \vert} n$ Thus, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent: :$\left \vert{1 + \dfrac z n}\right \vert^n \le \left({1 + \dfrac {\left \vert{z}\right \vert} n}\right)^n$ Call this result $(1)$. Also: {{begin-eqn}} {{eqn | l = \left \vert{z}\right \vert       | o = <       | r = M' }} {{eqn | ll= \implies       | l = 1 + \frac {\left \vert{z}\right \vert} n       | o = <       | r = 1 + \frac {M'} n       | c = Divide both sides by $n$, add $1$ to both sides }} {{eqn | n = 2       | ll= \implies        | l = \left({1 + \frac {\left \vert{z}\right \vert} n}\right)^n       | o = <       | r = \left({1 + \frac {M'} n}\right)^n       | c = Power Function is Strictly Increasing over Positive Reals: Natural Exponent }} {{end-eqn}} Thus it is sufficient to show that $\left({1 + \dfrac {M'} n}\right)^n$ is bounded. From Exponential Sequence is Eventually Increasing: :$\exists N \in \N: n \geq N \implies  \left({1 + \dfrac {M'} n}\right)^n \le \left({1 + \dfrac {M'} {n + 1}}\right)^{n + 1}$ So, for $n \geq N$: {{begin-eqn}} {{eqn | l = \left({1 + \frac {M'} n}\right)^n       | o = \leq        | r = \lim_{n \mathop \to \infty} \left({1 + \frac {M'} n}\right)^n       | c = Limit of Bounded Convergent Sequence is Bounded }} {{eqn | r = \exp M' }} {{end-eqn}} Now, for each $n \in \left \{{1, 2, \ldots, N-1}\right \}$: :$\exists M_n \in \R: \left({1 + \dfrac {M'} n}\right)^n \leq M_n$ by Continuous Function on Compact Subspace of Euclidean Space is Bounded. So: :$\forall n \in \N: \left({1 + \dfrac {M'} n}\right)^n \le \max \left \{{M_1, M_2, \ldots, M_{N-1}, \exp M'}\right \}$ Call this result $(3)$. Hence, for $z \in U$; that is, for $\left \vert{z}\right \vert < {M'}$: {{begin-eqn}} {{eqn | l = \forall n \in \N: \left \vert{\left({1 + \frac z n}\right)^n}\right \vert       | r = \left \vert{1 + \dfrac z n}\right \vert^n       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | o = \le       | r = \left({1 + \dfrac {\left \vert{z}\right \vert} n}\right)^n       | c = From $(1)$ }} {{eqn | o = \le       | r = \left({1 + \frac {M'} n}\right)^n       | c = From $(2)$ }} {{eqn | o = \le       | r = \max \left \{ {M_1, M_2, \ldots, M_{N-1}, \exp M'}\right \}       | c = From $(3)$ }} {{end-eqn}} That is, $\mathcal E$ is uniformly bounded on $U$. === $\mathcal E$ is Uniformly Convergent on $K$ === Note that, from Combination Theorem for Complex Derivatives: :$\forall n \in \N: E_n \left({z}\right)$ is holomorphic on $U$. From Uniformly Bounded Family is Locally Bounded, $\mathcal E$ is locally bounded on $U$. From Montel's Theorem, $\mathcal E$ is a normal family. From Vitali's Convergence Theorem, $\mathcal E$ converges uniformly on any compact subset of $U$. In particular, $\mathcal E$ converges uniformly on $K$. Hence the result. {{qed}} Category:Exponential Function nt96mtpfcinreoq2gfzfitnm9x5wqg5