Derivative: $e^x$. 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 $\ln(x)$.
 A: We must calculate the limit
$$
\lim_{h \to 0} \frac{e^h -1}{h}
$$
using the definition we have for $e$ we have
$$
\lim_{h \to 0}\lim_{n \to \infty}\frac{(1+1/n)^{hn} -1}{h}
$$
we expand using the binomial theorem:
$$
(1+ 1/n)^{hn} = \sum_{k=0}^{\infty}\binom{hn}{k}(1/n)^k = 1 + h + h^2 \cdots
$$
Then the tricky part (this needs to be justified carefully), we exchange the limits to get:
$$
\lim_{n \to \infty} \lim_{h\to 0}\frac{1 + h + h^2 \cdots -1}{h} = \lim_{n \to \infty} \lim_{h\to 0}(1 + h\cdots) = \lim_{n \to \infty}1 =1
$$
A: Bernoulli's inequality gives that for any $n\geq 2$ we have:
$$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n-1}\right)^n\tag{1} $$
hence $n\left(e^{\frac{1}{n}}-1\right)$ is between:
$$ 1 < n\left(e^{\frac{1}{n}}-1\right) < 1+\frac{1}{n-1}\tag{2}$$
hence, by squeezing:
$$ \lim_{n\to +\infty}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}=1 \tag{3}$$
that, together with $(2)$, gives:
$$ \lim_{r \to 0}\frac{e^r-1}{r}=1\tag{4}$$
that is enough to grant:
$$ \lim_{h\to 0}\frac{e^{x+h}-e^{x}}{h}=e^x \tag{5} $$
as wanted.
A: 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:
$$
\left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{4}
$$
A: \begin{equation*}
 e^x=\lim_{n\rightarrow \infty}\left(1+\frac {x} {n}\right)^n =\lim_{n\rightarrow \infty}\sum_{k=0}^n\binom n k\frac {x^k} {n^k}
=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {k!}\frac {x^k} {n^k}\end{equation*}
$$
=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {n^k}\frac {x^k} {k!}=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n} {n}\frac {n-1}{n}...\frac{n-k+1} {n}\frac {x^k} {k!}$$
$$\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {1} {1}\left(1-\frac {1}{n}\right)...\left(1-\frac{k-1} {n}\right)\frac {x^k} {k!}=\sum_{k=0}^\infty\frac {x^k} {k!}$$
Hence we have that
$$e^x=\sum_{k=0}^\infty\frac {x^k} {k!}\implies(e^x)'=\sum_{k=1}^\infty\frac{kx^{k-1}} {k!}=\sum_{k=1}^\infty\frac{x^{k-1}} {(k-1)!}=\sum_{k=0}^\infty\frac{x^{k}} {k!}$$
It follows that $e^x=(e^x)'$
A: Define the log function as $\log x=\int_1^x \frac{dt}{t}$.  The derivative is obviously $1/x$.  The exponential function is the inverse of the log function.  So, we may write
$$\begin{align}
x&=\log (e^x)\\
&=\int_1^{e^x} \frac{dt}{t}
\end{align}$$
Taking the derivative of both sides and applying the chain rule reveals
$$\begin{align}
\frac{dx}{dx}&=1\\
&=\frac{d\log(e^x)}{dx}\\
&=\frac{1}{e^x}\frac{de^x}{dx}
\end{align}$$
whereupon solving for $\frac{de^x}{dx}$ shows that
$$\frac{de^x}{dx}=e^x$$
So, the derivative of the exponential function is  itself!
A: $$e^x=\sum_{n\ge 0} \frac{x^n}{n!}$$
$$\frac{de^x}{dx}=\frac{d}{dx}[1+x+x^2/2!+x^3/3!+\cdots]=0+1+x+x^2/2+\cdots=\sum_{n\ge 0} \frac{x^n}{n!}=e^x$$
This uses a theorem justifying interchanging the derivative with the summation.
