It is important to appreciate the behavior of exponential functions as
the input to them becomes a large positive number, or a large negative
number. This behavior is different from the behavior of polynomials or
rational functions, which behave similarly for large inputs regardless
of whether the input is large *positive* or large *negative*. By contrast, for exponential functions, the behavior is
radically different for large *positive* or large *negative*.

As a reminder and an explanation, let's remember that
exponential notation started out simply as an **abbreviation**: for
positive integer $n$,
$$2^n = 2 \times 2 \times 2\times \ldots \times 2\;\;\;\;\;\hbox{ ($n$
factors) }$$
$$10^n = 10 \times 10 \times 10\times \ldots \times 10\;\;\;\;\;\hbox{ ($n$
factors) }$$
$$\left({1\over 2}\right)^n = \left({1\over 2}\right) \times
\left({1\over 2}\right) \times \left({1\over 2}\right)\times \ldots
\times \left({1\over 2}\right) \;\;\;\;\;\hbox{ ($n$
factors) }$$

From this idea it's not hard to understand the **fundamental
properties of exponents** (they're not *laws* at all):
$$a^{m+n} = \underbrace{a\times a\times a\times \ldots \times a}_{m+n}\;\;\;\;\;\hbox{
($m+n$ factors)}$$
$$ = \underbrace{(a\times a\times a\times \ldots \times a)}_{m} \times
\underbrace{(a\times a\times a\times \ldots \times a)}_{n}
= a^m \times a^n$$
and also
$$a^{mn} = \underbrace{(a\times a\times a\times \ldots \times a)}_{mn}
= $$
$$ = \underbrace{
\underbrace{(a\times a\times a\times \ldots \times a)}_{m} \times
\ldots \times
\underbrace{(a\times a\times a\times \ldots \times a)}_{m}
}_n = (a^m)^n$$
at least for positive integers $m,n$. Even though we can only easily
see that these properties are true when the exponents are positive
integers, the *extended* notation is guaranteed (by its *meaning*, not by *law*) to follow the same rules.

Use of *other* numbers in the exponent is something that
came later, and is also just an *abbreviation*, which happily
was *arranged* to match the more intuitive simpler version. For
example,
$$a^{-1} = {1\over a}$$
and (as consequences)
$$a^{-n} = a^{n\times(-1)} = (a^n)^{-1} = {1\over a^n}$$
(whether $n$ is positive or not). Just to check one example of
consistency with the properties above, notice that
$$a = a^1 = a^{(-1)\times (-1)} = {1\over a^{-1}} = {1\over 1/a} = a$$
This is not supposed to be surprising, but rather reassuring that we
won't reach false conclusions by such manipulations.

Also, fractional exponents fit into this scheme. For example
$$a^{1/2} = \sqrt{a}\;\;\;\;\;a^{1/3} = \sqrt[3]{a}$$
$$a^{1/4} = \sqrt[4]{a}\;\;\;\;\;a^{1/5} = \sqrt[5]{a}$$
This is *consistent* with earlier notation: the fundamental
property of the $n^{\rm th}$ root of a number is that its $n^{\rm th}$
power is the original number. We can check:
$$a = a^1 = (a^{1/n})^n = a$$
Again, this is not supposed to be a surprise, but rather a consistency
check.

Then for arbitrary *rational* exponents $m/n$ we can
maintain the same properties: first, the definition is just
$$a^{m/n} = (\sqrt[n]{a})^m$$

One hazard is that, if we want to have only real numbers (as
opposed to complex numbers) come up, then we should not try to take
square roots, $4^{\rm th}$ roots, $6^{\rm th}$ roots, or any *even* order root of negative numbers.

For general *real* exponents $x$ we likewise should *not* try
to understand $a^x$ except for $a>0$ or we'll have to use complex
numbers (which wouldn't be so terrible). But the value of $a^x$ can
only be defined as a *limit*: let $r_1,r_2,\ldots$ be a sequence
of *rational* numbers approaching $x$, and define
$$a^x = \lim_i \;a^{r_i}$$
We would have to check that this definition does not accidentally
depend upon the sequence approaching $x$ (it doesn't), and that the same properties
still work (they do).

The number $e$ is not something that would come up in really elementary mathematics, because its reason for existence is not really elementary. Anyway, it's approximately $$e = 2.71828182845905$$ but if this ever really mattered you'd have a calculator at your side, hopefully.

With the definitions in mind it is easier to make sense of
questions about **limits** of exponential functions. The two
companion issues are to evaluate
$$\lim_{x\rightarrow +\infty}\;a^x$$
$$\lim_{x\rightarrow -\infty}\;a^x$$
Since we are allowing the exponent $x$ to be *real*, we'd better
demand that $a$ be a *positive real* number (if we want to avoid
complex numbers, anyway). Then
$$\lim_{x\rightarrow +\infty}\;a^x =
\left\{\matrix{
+\infty& \hbox{ if } & a>1 \cr
1& \hbox{ if } & a=1 \cr
0& \hbox{ if } & 0< a<1
}\right.$$
$$\lim_{x\rightarrow -\infty}\;a^x =
\left\{\matrix{
0& \hbox{ if } & a>1 \cr
1& \hbox{ if } & a=1 \cr
+\infty& \hbox{ if } & 0< a<1
}\right.$$

To remember which is which, it is sufficient to use $2$ for $a>1$ and ${1\over 2}$ for $0< a<1$, and just let $x$ run through positive integers as it goes to $+\infty$. Likewise, it is sufficient to use $2$ for $a>1$ and ${1\over 2}$ for $0< a<1$, and just let $x$ run through negative integers as it goes to $-\infty$.