We present an elementary proof that:

At the outset, we observe that $\sum_{k=0}^\infty z^k/k!$ converges by the ratio test. For definiteness, the notation $e^z$ below will refer to exactly this series.

Proof. We expand the right-hand side in the straightforward manner:

where $\pi(k, n)$ denotes the coefficient

Let $\abs{z} \leq M$ . Given $\epsilon > 0$ , there is a $N \in \nat$ such that whenever $n \geq N$ , then $\sum_{k=n+1}^\infty M^k/k! < \epsilon/2$ , since the sum is the tail of the convergent series $e^M$ .

Since $\lim_{n \to \infty} \pi(k,n) = 1$ for fixed $k$ , there is also a $N' \in \nat$ , with $N' \geq N$ , so that whenever $n \geq N'$ and $0 \leq k \leq N$ , then $\abs{\pi(k, n) - 1} < \epsilon/(2e^{M})$ . (Note that $k$ is chosen only from a finite set.)

Now, when $n \geq N'$ , we have

(In the middle sum, we use the bound for all and .)

In fact, we have proved uniform convergence of $\lim_{n\to\infty} \left( 1 + \frac{z}{n} \right)^n$ over $\abs{z} \leq M$ . Exploiting this fact we can also show:

Proof. Fix $\abs{z} < M$ . Given $\epsilon > 0$ , for large enough $n$ , we have

Since $o(1) \to 0$ , for large enough $n$ we can set $w = z+o(1)$ above. Since the exponential is continuous ¹, for large enough $n$ we also have $\abs{e^{z+o(1)} - e^z} < \epsilon/2$ . Thus

Footnotes

...http://planetmath.org/encyclopedia/ContinuousMap.html ¹

follows from uniform convergence on bounded subsets of either expression for $e^z$