I'm looking for a straight forward proof using the definition of a derivative applied to the exponential function and substitution of one of the limit definitions of $e$, starting with
$e = \lim_{h\to \infty}\left({1+\dfrac{1}{h}}\right)^h$ or $e=\sum_{h=0}^{\infty}{\dfrac{1}{h!}}$
and
$\dfrac{d}{dx}\left( e^x \right) = \lim_{h\to 0}\left({\dfrac{e^{x+h}-e^{x}}{h}}\right)$
I found a proof I sort of liked here (which is sort of along the lines of a proof I'd like to use):
http://www.math.brown.edu/UTRA/explog.html
My only problem is that he combines the dummy variable, $h$, for the limit definition of $e$ and the dummy variable, $h$, used for the derivative. To me, it seems like it's not quite valid to do such a thing because it assumes both values are equal. Can anyone provide a better proof or justification for why the dummy variables can be combined?
EDIT:
I guess I'd also like to have a proof of why:
$\lim_{h\to 0}\left({\dfrac{e^{h}-1}{h}}\right) = 1$
using one of the limit definitions of $e$ shown above.
$$\frac{e^h-1}h=\frac1h\left(h+\frac{h^2}{2!}+\frac{h^3}{3!}+\ldots\right)=1+\frac h{2!}+\frac{h^2}{3!}+\ldots\xrightarrow[h\to 0]{}1$$
Of course, some power series theory must be known to fully justify the above. And now all it's easy:
$$\lim_{h\to 0}\frac{e^{x+h}-e^x}h=\lim_{h\to 0}\,e^x\frac{e^h-1}h=e^x\cdot 1=e^x$$