alternate interpretation on the proof of $\mathrm{d}(e^x) = e^x$ I'm trying to develop a "natural sense" of calculus rather than just remembering formulas. 
The proof of $de^x=e^x$ is rather interesting and I wonder if step (3) below means anything special? Is there an alternate, perhaps geometric, interpretation for it? Does it bear any special insight into why $e^x$ is such an important function in calculus?
(1) starting from the definition:
$de^x = \lim_{\Delta x \to 0} \cfrac{e^{x+\Delta x} - e^x}{\Delta x}$
(2) after some algebra, I get: $d(e^x) = \lim_{\Delta x \to 0} \cfrac{e^x (e^{\Delta x} - 1)}{\Delta x}$
I noticed that $\lim_{\Delta x \to 0} \cfrac{e^{\Delta x} - 1}{\Delta x}$ is the the slope of $e^x$ when $x = 0$. so I cheat a little without going all the way to the definition of $e$ using limits.
(3) rewrite as $de^x = \lim_{\Delta x \to 0} e^x * \lim_{\Delta x \to 0} \cfrac{e^{\Delta x} - 1}{\Delta x}$
It seems to suggest that the derivative of $e^x$ is itself multiplied by the slope of itself when $x=0$ but saying so makes little sense. why is that? what am I missing?
 A: You are completely right!
It turns out that $e^x$ is the only continous and differentiable function such that $f^\prime=f$ and $f(0)=1$. If we omit the last condition, then we get the class of all exponential functions - namely all functions $ae^x$. You will see that you get the same calculations if you use $a^x$ instead of $e^x$. What is special about $e$ is precisely that the limit
 $$\lim_{h \to 0} \frac{{e^h}-1}{h}$$
is equal to one.
A: For any exponential function $a^x$, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}a^x
&=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}\\
&=a^x\lim_{h\to0}\frac{a^h-1}{h}\\[6pt]
&=c_a\,a^x
\end{align}
$$
Now, assuming the interchange of limits is okay,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}e^x
&=\frac{\mathrm{d}}{\mathrm{d}x}\lim_{n\to\infty}\left(1+\frac xn\right)^n\\
&=\lim_{n\to\infty}\frac nn\left(1+\frac xn\right)^{n-1}\\
&=e^x\lim_{n\to\infty}\left(1+\frac xn\right)^{-1}\\[6pt]
&=e^x
\end{align}
$$
Therefore, $c_e=1$. In fact, the chain rule yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}a^x
&=\frac{\mathrm{d}}{\mathrm{d}x}e^{x\log(a)}\\
&=\log(a)\,e^{x\log(a)}\\[6pt]
&=\log(a)\,a^x
\end{align}
$$
Therefore, $c_a=\log(a)$.
A: "Does it bear any special insight into why $e^x$ is such an important function in calculus?"  Yes.  If you think of $\frac{d}{dx}$ as a linear operator that maps $f(x)$ to $f'(x)$, then it is reasonable to ask if such a linear operator has any invariant subspaces (i.e. $f'(x) = af(x)$) like matrices do.  For matrices we call these invariant subspaces eigenvectors, for a linear operator like $\frac{d}{dx}$ we would call it an eigenfunction.  Why this is so important is that in solving real-world problems? We often encounter (in physics and engineering) equations where an unknown function is a linear combination of its own derivatives.  For example, 
$$
f''(x) + af'(x) + bf(x) = 0
$$
It is clear that in order to solve it we need to find a function that does not fundamentally change when taking the derivative.  It turns out that $e^{cx}$ is just that function.  That is why $\frac{d}{dx}e^x = e^x$ is so important.   It helps us solve practical problems.
