TITLE: Derivative: $e^x$.
QUESTION [9 upvotes]: How do you differentiate $e^x$?
I looked on many sites, including similar questions here but most answers seemed circular.
The only known definition of $e$ to be used in this proof is 
$$
e=\lim_{n \to\infty} \left(1+\frac{1}n \right)^n
$$
What I did is:
$$
\begin{align*}
(e^x)' &=\lim_{h\to0}\frac{e^{x+h}-e^x}h \\
&= e^x\lim_{h\to0}\frac{e^{h}-1}h
\end{align*}
$$
But I don't know how to go on, I know  $\lim_{h\to0}\frac{e^{h}-1}h=1$ but I don't know how to prove it, I can't use the $e^x$ taylor expansion as that would imply diferentiating $e^x$.
Edit: I also can't use the derivative of $\ln(x)$.

REPLY [6 votes]: This is really a comment on Jack D'Aurizio's answer, but it is too long for a comment. Let me show how Bernoulli's Inequality is used to prove $(1)$ from Jack D'Aurizio's answer.
In this answer, it is shown, using Bernoulli's Inequality, that
$$
\left(1+\frac1n\right)^n\tag{1}
$$
is an increasing sequence and that
$$
\left(1+\frac1n\right)^{n+1}\tag{2}
$$
is a decreasing sequence. This means that for all $n$
$$
\left(1+\frac1n\right)^n\le\overbrace{\lim_{k\to\infty}\left(1+\frac1k\right)^k}^{\large e}\le\left(1+\frac1n\right)^{n+1}\tag{3}
$$
Since $(3)$ is true for all $n$, we can substitute $n\mapsto n-1$ in the right-side inequality to get the inequality from Jack's answer:
$$
\left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{4}
$$