Okay, I'll be a little stubborn here and I'll try to convince you that defining $e^x$ as a power series is actually better. (Respecting your second wish, I'll not enter in a lot of technicalities)
First, let me say something: People usually think that mathematicians do things in order to complicate simple things. The fact is that most of times, the mathematician is actually simplifying things. As a rule of thumb, in order to simplify something, you need to make a theory, and the more you want to simplify, the more abstract or hard to grasp the theory will be.
Now, okay, let's make a function called $\xi (x)$, defined by:
$\displaystyle \xi (x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ (Here, we adopt $0^0=1$ to avoid notational complications)
Now, don't be afraid of this definition (I didn't even say that this is the exponential! This is a auxiliary function, for now)
The important things are:
- $\xi$ is continuous. (This follows from it being a power series)
- $\xi (x+y) = \xi(x)\xi(y)$. (This follows from the Cauchy Product Formula)
- $\xi$ is defined for ALL reals. (This is a simple observation, but will be extremely useful)
- $\xi(0)=1$, $\xi(1)=e$
And now, let's take a look at the function you call "exponential". Let's call it $\exp$. We have:
$\exp(x+y)=\exp(x)\exp(y)$ for all $n \in \mathbb{N}$ (I hope this is clear)
$\exp(0)=1$, $\exp(1)=e$
Now, for $n \in \mathbb{N}$, we have that $\exp(n)=\exp(1)...\exp(1)$ $n$ times=$\exp(1)^n$
For $-n$ negative integer, we have that $1=\exp(0)=\exp(n-n)=\exp(n)\exp(-n)$, hence $\displaystyle \exp(-n)=\frac{1}{\exp(n)}=\frac{1}{\exp(1)^n}=\exp(1)^{-n}$
For $a=\frac{p}{q} \in \mathbb{Q}$, $\exp(p)=\exp(q.\frac{p}{q})=\exp(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $q$ times $=\exp(p/q)^q$. Hence, $\exp(\frac{p}{q})=\exp(p)^{1/q}=(\exp(1)^p)^{1/q}$
Now, note the following:
- I computed the values of $\exp(x)$ for all $x$ rational using only properties that $\xi$ AND $\exp$ have in common. Hence, $\xi$ coincides with $\exp$ in the rationals.
- By the previous observation, we have immediately that $\exp$ is continuous in the rationals, as a restriction of a continuous function is continuous.
Okay, now, since we have that $\xi$ is defined for ALL reals, we could very well say:
Well, make $\xi$ be what we will call $\exp$ for the reals, then.
But we can do more. Since $\exp$ coincides with $\xi$ in a dense set ($\mathbb{Q}$), if we want it to extend continuously, the ONLY way is for the extension to be $\xi$ (maybe this is not clear for you, but it is a theorem, that is stated as follows, and holds in a way more general situation:)
Two continuous functions that are equal in a dense subset of $\mathbb{R}$ are equal at the whole domain.
Now, notice the irony. You began asking why $\exp$ is continuous, and we derived quite "easily" that $\exp$ is continuous in the rationals, and the motivation came as to whether we could have a continuous "$\exp$" in the reals.
Just to close up, you may feel like I'm cheating in requiring a continuous extension. Well, I'm not. Following your intuition, I'd say that the only things that you consider the exponential surely satisfies is $\exp(x+y)=\exp(x).\exp(y)$, $\exp(0)=1$ and $\exp(1)=e$. Well, there are a LOT of non-continuous functions satisfying this in $\mathbb{R}$, coinciding with the values of $\exp$ in the rationals. And, by the way, your way of defining $\exp$ for reals as limits of sequences is blalantly, explictly requiring it to be continuous in $\mathbb{R}$.
EDIT: Appendix: Why considering THIS series:
Whatever $\exp$ will be, we want that $\exp(x+y)=\exp(x).\exp(y)$
Now, let's say we should guess a power series for $\exp(x)$
$$\exp(x)=\sum_{n=0}^{\infty}a_nx^n$$
Well, $\displaystyle \exp(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the binomial theorem.
But, hey, look! The right hand side is begging for the cauchy product formula. If that $a_n.n!$ were $1$... then, the right hand side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. So, why not? Let $a_n.n!$ be $1$, then! So, we have $\displaystyle a_n=\frac{1}{n!}$. Now, look back at the series for $\exp$, and perceive our luck.