Proofing that the exponential function is continuous in every $x_{0}$ Given:
$$\exp: \mathbb{R} \ni  x \mapsto \sum_{k=0}^{\infty } \frac{1}{k!} x^{k} \in \mathbb{R}$$
also $e = \exp(1)$. For all $x \in \mathbb{R}$ with $\left | x \right | \leq 1$: 
$$\left | \exp(x) - 1 \right | \leq \left | x \right | \cdot (e-1)$$
and $\exp(0) = 1$
...........................................................................................................................
In order to proof that the exponential function is continuous for every $x_{0}$,
it needs to be shown that it's continuous at all.
This was shown here (it's continuous at $x_{0}$ = 0): Proving that the exponential function is continuous
But I prefer this proof:
$$
\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}
$$
apply some little changes
$$
\exp(x) = 1 + x \sum_{n=1}^\infty \frac{x^{n-1}}{n!}
$$
whence for $x\to 0$
$$
|\exp(x) - 1| \le |x| \sum_{n=1}^\infty{|x|^{n-1}} \le |x| \frac{1}{1-|x|} \to 0
$$
I would say in order to show that the exponential function is continuous for all $x_{0} = 0$, I just need to show it is continuous at $x_{0}$ = 0 (done) and then I can just conclude it is continuous everywhere, so at $x=x_{0}$?
Not sure about this, is it really possible?
 A: The proof that
$$
\exp(a+b)=\exp(a)\exp(b)
$$
is a simple application of absolute convergence and binomial theorem. See Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series
Once you have this knowledge, you can observe that
$$
|\exp(x+h)-\exp(x)|=|\exp(x)|\,|\exp(h)-1|
$$
and use the already proved continuity at $0$.
A: 
I thought it might be instructive to show that $e^xe^y=e^{x+y}$ directly from the limit definition of the exponential function given by
$$\bbox[5px,border:2px solid #C0A000]{e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n} \tag 1$$
To that end, we now proceed


First, we see that 
$$\begin{align}
\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n&=\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n\\\\
&=\left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n \tag 2
\end{align}$$
If $xy>0$, then for all $n>N$ we have from $(2)$
$$\left(1+\frac{x+y}{n}\right)^n\le  \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n \le \left(1+\frac{x+y+\frac{xy}{N}}{n}\right)^n \tag 3$$
Taking the limit as $n\to \infty$ of $(3)$ reveals
$$e^{x+y}\le e^{x}e^{y} \le e^{x+y+xy/N} \tag 4$$
Using the left-hand side inequality in $(4)$ and applying it to the right-hand side shows that for $xy>0$, $e^{x+y+xy/N}\le e^{x+y}e^{xy/N}$ and therefore
$$e^{x+y}\le e^{x}e^{y} \le e^{x+y} e^{xy/N} \tag 5$$
Letting $N\to \infty$ yields
$$e^{x+y}\le e^xe^y\le e^{x+y}$$
where we used the inequalities for the exponential $1+x\le e^x\le \frac{1}{1-x}$ for $x<1$ that I established in THIS ANSWER using only $(1)$ and Bernoulli's Inequality.
We have now established the equality $e^xe^y=e^{x+y}$ for $xy>0$.  For $xy<0$, we simply reverse the inequalities in $(3)$ and proceed analogously.

Finally, refer to THIS ANSWER in which the result herein was used to prove the continuity of $e^x$ for all $x$.

