Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:
$$\begin{align}
(e^x)' &= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\
&= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\
&= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\
&= e^x \lim_{h \to 0}\frac{e^h-1}{h}
\end{align}
$$
I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint?

1) Using Apostol's definition in "Calculus I", 6.12, or in Finney's " Calculus and Analytic Geometry". 6.3  , or in Swokowski's "Calculus and Analytic Geometry", def. 7.8, we have that $$\,\,e^x=y\iff^{\textrm{def.}} \log y=x$$  and assuming we know the usual about the logarithmic function we get $$y=e^x\implies \log y=x\implies\frac{1}{y}\,dy=dx\implies \frac{dy}{dx}=y=e^x\implies (e^x)'=e^x$$ and now using the definition of derivative, for any real we have $$e^x=(e^x)'=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\to 0}\frac{e^h-1}{h}\implies \lim_{h\to 0}\frac{e^h-1}{h}=1$$since, by this def. of the exponential function, we get using the basic properties of logarithm $$e^{x+h}=e^xe^h\iff \log(e^xe^h)=x+h ....$$$${}$$
(2) By one of the most usual, and elementary, of the definitions of the number $\,e\,$ , we get: $$e:=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\implies e^h=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{hn}=\lim_{k\to\infty}\left(1+\frac{h}{k}\right)^k$$putting $\,\,k:=hn\,\,$, so$$\lim_{h\to 0}\frac{e^h-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\left(1+\frac{h}{n}\right)^n-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\left(\frac{h}{n}\right)^k-1}{h}=$$$$=\lim_{h\to 0}\lim_{n\to\infty}\frac{\sum_{k=1}^n\binom{n}{k}\left(\frac{h}{n}\right)^k}{h}=1+\lim_{h\to 0}\lim_{n\to\infty}\rlap{/}{h}\frac{\sum_{k=2}^n\binom{n}{k}\frac{h^{k-1}}{n^k}}{\rlap{/}{h}}=1$$as every single summand in what's left in that sum is multiplied by a positive power of $\,h$...
As far as I know, most authors go with the first approach, I couldn't find even one that goes with the second approach, and some even go with the definition of the exponential function as a power series...now you choose!
