TITLE: The derivative of $e^x$ using the definition of derivative as a limit and the definition $e^x = \lim_{n\to\infty}(1+x/n)^n$, without L'Hôpital's rule
QUESTION [12 upvotes]: Let's define 
$$
e^x := \lim_{n\to\infty}\left(1+\frac{x} {n}\right)^n, \forall x\in\Bbb R
$$
and
$$
\frac{d} {dx} f(x) := \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)} {\Delta x}
$$
Prove that
$$
\frac{d} {dx} e^x = e^x
$$
using the definition of $e^x$ and derivation above, without using L'Hôpital's rule or the "logarithm trick" and/or the "inverse function derivative trick".
$$
\left( \frac{d} {dx} f^{-1}(x)= \frac{1} {\left(\frac{d}{d(f^{-1}(x))}f(x)\right)(f^{-1}(x))}\right)
$$
Or equivalently prove that the following two definiton of $e$ are identical
$$
1)\space\space\space\space e =\lim_{n\to\infty}(1+\frac{1} {n})^n
$$
$$
2) \space\space\space\space e\in\Bbb R,\space\space(\frac{d} {dx} e^x)(x=0) = 1  
$$
What I've got is
$$
\frac{d}{dx}e^x=e^x \lim_{\Delta x\to0}\frac{e^{\Delta x} - 1} {\Delta x} =
e^x \lim_{\Delta x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{\Delta x}{n}\right)^{n}-1}{\Delta x} = 
e^x \lim_{\Delta x\to0}\frac{e^{0+\Delta x}-e^0}{\Delta x} 
$$
If i assume that $n\in\Bbb N$ I could use binomial theorem but I didn't got much out of it. Wolframalpha just uses L'Hospital rule to solve it, but I am looking for an elementary solution. What I'm interested in is basically is the equivalence of the two definition of $e$ mentioned above. And I'd like to get a direct proof rather than an indirect(I mean which involves logarithms or the derivatives of invers functions).
I look forward getting your aswers.

REPLY [8 votes]: First, as in Paramanand Singh's answer, let's prove that $e^{x+y}=e^xe^y$. To do that, note that, for big $n$, from Bernoulli's inequality, $$\begin{align}\frac{\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n}{\left(1+\frac{x+y}{n}\right)^n}&=\left(\frac{1+\frac{x+y}{n}+\frac{xy}{n^2}}{1+\frac{x+y}{n}}\right)^n\\
&=\left(1+\frac{xy}{n^2+nx+ny}\right)^n\\
&\geq1+\frac{nxy}{n^2+nx+ny},\end{align}$$ which goes to $1$ as $n\to \infty$, and also 
$$\begin{align}\frac{\left(1+\frac{x+y}{n}\right)^n}{\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n}&=\left(\frac{1+\frac{x+y}{n}}{1+\frac{x+y}{n}+\frac{xy}{n^2}}\right)^n\\
&=\left(1-\frac{xy}{n^2+nx+ny+xy}\right)^n\\
&\geq1-\frac{nxy}{n^2+nx+ny+xy},\end{align}$$ which also goes to $1$ as $n\to \infty$. So, the ratio coverges to $1$, therefore $e^{x+y}=e^xe^y$.
Now, if $\displaystyle e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$, then, using Bernoulli's inequality, you have that $\displaystyle\frac{x}{n}>-1$ for big $n$, so, for those $n$, $$\left(1+\frac{x}{n}\right)^n\geq 1+n\frac{x}{n}=1+x\Rightarrow e^x\geq 1+x.$$ This holds for all $x\in\mathbb R$, so it also holds for $-x$, therefore, for $x<1$, $$e^{-x}\geq 1-x\Rightarrow e^x\leq\frac{1}{1-x}\Rightarrow e^x-1\leq\frac{x}{x-1},$$ where we used that $e^xe^{-x}=e^0=1$. So, if $x\in(0,1)$, from the second inequality you have that $\displaystyle\frac{e^x-1}{x}\leq\frac{1}{1-x}$, and the first inequality gives that $\displaystyle\frac{e^x-1}{x}\geq 1$. So, $\displaystyle\lim_{x\to 0^+}\frac{e^x-1}{x}=1$. You treat the case $x<0$ similarly, and you have that $e^x$ is differentiable at $0$, with derivative equal to $1$.
Using that, you have that $$\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\to 0}\frac{e^h-1}{h}=e^x,$$ so the derivative of $e^x$ is $e^x$.