Could you explain why $\frac{d}{dx} e^x = e^x$ "intuitively"?
Well, think of exponential growth (like e.g. bacteria grow):
We know, the more bacteria exist in a colony, the faster the colony will grow. More precisely: The growth speed of the colony $B$ is proportional to it's size ... Double size, double speed.
$$\frac{dB}{dt} \sim B$$
Furthermore we know the growth is exponential, since bacteria clone themselves in fixed amounts of time, i.e.
$$B \sim 2^{k\cdot t}$$
Putting it together, we can deduce that:
$$\frac{d}{dt}2^{kt} = c \cdot 2^{kt}$$
or with $a = 2^k$
$$\frac{d}{dt}a^t = c \cdot a^t$$
Now, how do we get the $e$? We just ask: What base $a$ do we have to take such that $c = 1$, i.e. $\dot{B} = B$?
We simply call that base $e$. Having such an $e$ is quite useful. We could use its special derivation traits we found above to define all exponential functions to the base $e$.
$$a^x = e^{x \cdot \ln a} $$
This shows that the factor $c$ we encountered in the above equations equals $\ln a = \log_e a$ and therefore, we can easily derive all kinds of exponential terms.
After all, $e$ turns out to be $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \approx 2.718\ldots$
(Looks like Niel and "J" beat me to the punch while I was composing this, but I'll post my answer anyway since I like my diagram. :) )
You can't get too much more intuitive than the Law of Exponents ($a^x a^c = a^{x+c}$) right?
To reduce symbolic (and mental) clutter for a while, let's write $k$ for $a^c$, to get
$$k \; a^x = a^{x+c}$$
The point is that
Multiplying the value of $a^x$ by something yields the same result as adding some other thing to the exponent.
Consider what this bit of algebra tells us about the geometry of the graph of the function $y=a^x$ (shown in red in the figure below). Recall some fundamental notions:
$y=k a^x$ is the result of vertically stretching (scaling) the original graph by a factor of $k$.
$y = a^{x+c}$ is the result of horizontally sliding (translating) the original graph by a (signed) distance of $-c$.
The Law of Exponents tells us that (provided $k$ and $c$ are related appropriately), these transformed graphs are identical! In the figure, the blue graph represents both results.[*] Importantly, what points move where aren't the same under both actions; scaling moves the red point $P$ vertically onto the blue point $Q$; translating moves the red point $R$ horizontally onto $Q$.
Now, as suggested by the diagram, imagine a tangent vector poking out of each point of the original graph, and consider what the transformations to do those vectors:
- Vertically stretching scales (only) the "rise" of the vector by factor $k$; that is, a vector with slope $m$ is pulled to achieve a slope of $m k$.
- Horizontally translating has no effect on the vector's slope.
What can we conclude here? Why, something pretty remarkable:
For any point $P$ on the original graph, the point $R$ --located $c$ units to the right[**] of $P$-- is such that the slope of the tangent vector at $R$ is $k$-times the slope of the tangent vector at $P$ (where $k$ and $c$ are related appropriately).
Let me take this opportunity to retire $k$, since it is beginning to become clutter; I'll just make the appropriate relationship explicit. Also, I'm going to retire the point name $R$, opting to describe the point instead. I'll summarize things this way:
On the graph $y=a^x$, if the slope of tangent vector at any $P$ is $m$, then the slope of the tangent vector at the point $c$ units to the right of $P$ is $m\;a^c$.
Notice that there's nothing special about the players in this game. $P$ is any point on the graph, $c$ is any (horizontal) distance you care to choose; heck, even the exact value of $a$ is up for grabs. Let's use this to our advantage.
Suppose we take $P$ to be the point where the (original, red) graph crosses the $y$-axis; that is, we take $P$ to have $x$-coordinate $0$ (and $y$-coordinate $1$, but this doesn't really matter). Then, "the point $c$ units to the right of $P$" can be described more simply as "the point with $x$-coordinate $c$", and we have
On the graph $y = a^x$, if the slope of the tangent vector at the point with $x$-coordinate $0$ is $m$, then the slope of the tangent vector at the point with $x$-coordinate $c$ is $m\;a^c$.
In Calculusian prose:
If $f(x) = a^x$, then $f^{\prime}(c) = a^c f^{\prime}(0)$.
Observe that we don't even need "$c$" in the above formula, since it's just taking the place of some $x$-coordinate. We can simply write:
If $f(x) = a^x$, then $f^{\prime}(x) = a^x f^{\prime}(0)$.
From here, it's pretty much a matter of definition to get to the final answer to your specific question. After all, the above holds for any (non-negative) value of $a$. Clearly, some values of $a$ correspond to graphs that cross the $y$-axis very steeply; some values correspond to graphs that cross the $y$-axis very shallowly; it's not un-reasonable to believe that there's a convenient value would cause the graph to cross the $y$-axis juuuuuuuuust right ... with a slope of $1$. Of course (as I mix my folklore), to name a demon is to control him, so we'll simply assign a symbol to this "just right" value of $a$.
Let "e" be the number such that the tangent vector to $y=e^x$ at $x=0$ has slope $1$.
Assuming that such a number really does exist, you don't even have to know its exact value to conclude
If $f(x) = e^x$, then $f^{\prime}(x) = e^x$.
You can then turn your attention to figuring out why $e$ happens to have the value $2.718...$ . Other answers in this thread provide insights on how those arguments proceed.
Also left as an exercise is to determine why, in the formula for the derivative of $a^x$, the "$f^{\prime}(0)$" factor is in fact "$\log a$" (the natural logarithm of $a$).
[*] As Niel mentions, this illustrates that the graph of an exponential function is "self-similar". I think it's important to add "uni-directionally" to the description, in order to distinguish it from (conventional) similarity transformations where scaling occurs "omni-directionally".
[**] "to the left" works, too, with appropriate changes to the discussion.
Doesn't quite explain $e^x$, but looking at the series of $2^n$, i.e. $\{1,2,4,8,16,32,...,2^k,2^{k+1},...\}$ might help.
You see there if you take the first order difference you get back the same series, as $2^{k+1}-2^k=2^k$.