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.