You can prove this in several ways:
$(1)$ The exponential function is continuous. A function is continuous at ($x=x_0$) $\iff$ it is sequentially continuous (at $x=x_0$), thus the result follows.
$(2)$ The exponential function is differentiable over $\Bbb R$. Given a convergent sequence $\{a_n\}$, the sequence will be bounded over some interval $[a,b]$. Then we will have
$$|\exp(x_n)-\exp(x_0)|=\exp(\chi)|x_m-x_0|<M|x_m-x_0|$$ by the mean value theorem where $M$ is the maximum of $\exp$ (which is $ =\exp'$) over $[a,b]$.
Note that the exponential function has the superb property that being continuous at $x=0$ immediately means it is continuous at every other real. This is because of the functional equation $$\exp(x+y)=\exp(x)\exp(y)$$
Indeed, making $$\exp(x+h)-\exp(x)$$ small accounts to making $$\exp(x)(\exp(h)-1)$$
small. This can happen if $\exp(h)\to 1 $ when $h\to 0$ which is indeed true and means $\exp(h)=\exp(0)=1$ whence $\exp(h)$ is continuous at $x=0$. Thus, all you need is to show, as you're being asked, is that the exponential function is continuous at $x=0$, which means that if $|h|$ is small, so is $|e^h-1|$. My way would be to argue that since
$$1=\log e=\lim_{h\to 0}\frac{e^h-1}{h}$$ then
$$\lim_{h\to 0}]({e^h-1})=\lim_{h\to 0}h\frac{e^h-1}{h}=0\cdot 1=0$$
and continuity follows.
It is simply the definition of $e^z$, the complex exponential. What the author did before is showed that the series $\sum_{n=0}^\infty \frac{z^n}{n!}$ converge for any number $z\in \mathbb{C}$. If that wasn't the case then the definition of $e^z$ would make no sense.
Best Answer
This is equivalent to trying to prove that
$$ 1 = \sum_{k=0}^{\infty} \frac{k}{k!}.$$
You can see that these are equivalent by dividing both sides by $e^z$ (which is never zero). If $z=0$ clearly both sides are the same. If $z\neq 0$, you can divide both sides by $z$ so in either case, you can neglect $ze^z$ on both sides.
Note how similar this is to the exponential function: $e^z = \sum\limits_{n=0}^{\infty} \dfrac{z^n}{n!}$. If you differentiate both sides you get $$e^z = \sum\limits_{n=0}^{\infty} \dfrac{nz^{n-1}}{n!}.$$
What happens if you set $z=1$? Do you get your desired result?