Using definition of derivative to find $\frac{d}{dx} (e^x)$ -- circular reasoning? Using the definition of derivative, one can find that $\frac{d}{dx} (e^x) = e^x$ using the fact that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. Fine. 
However, the proofs that I have seen of that latter fact have all used the power series $e^x = 1 + x + x^2/2 + \dots$ which, to my knowledge, require all derivatives of $e^x$ to compute. So, is this proof circular? And if so, how can we correctly show that $\frac{d}{dx} (e^x) = e^x$ using the definition of a derivative?
 A: There are many, many different ways you can approach this.
(1) Define $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$. Then taking a derivative and passing the limit through (this needs to be justified but it can be done by showing uniform convergence) we can show that $\frac{d}{dx}e^x = e^x$. Alternatively, we can use the limit definition of a derivative and this definition to show this.
(2) Define $e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$. See Rene's answer for finding the bound $$\limsup_{h \to 0} \frac{e^h - 1}{h} \leq \frac{k+1}{k}$$ to see how we can compute the limit and hence the derivative.
(3) Define $e^x$ to be the number $y(x)$ such that $\int_1^{y(x)} \frac{1}{t} dt = x$. Use the fundamental theorem of calculus and the chain rule when taking a derivative of the above with respective to $x$ to get
$$\frac{1}{y(x)} y'(x) = 1$$
and conclude $y'(x) = y(x)$ or $(e^x)' = e^x$.

Now all these definitions are equivalent. That is, starting with one definition, we can show that the other must hold. It is clear that (2) and (3) imply (1) because knowing the derivative of $e^x$ you can find it's taylor series. But this then means that (2) and (3) give the same function. So they are all equivalent.
The truly circular way to compute the limit of the finite difference is to use L'Hospital's rule:
$$\lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{e^h}{1} = 1.$$
The problem is that you use what the derivative of $e^h$ is to show what the derivative of $e^x$ is.
A: Well I dont know if this is the answer you want, but here is a proof of 
$$\lim\limits_{x \to 0} \frac{e^{x}-1}{x}=1$$
direct from the definition of $e^x$ as limit. 
Note that $\left(1+\frac{x}{k} \right)^{k+1}$ is a decreasing sequence (induction, and Bernoulli inequality) and its limit as $k\to \infty$ is $e^x$ by the definition of $e^x$.
Now we have 
$$1\leq \frac{e^{x}-1}{x}
\leq \frac{1}{x}\left[\left( 1+\frac{x}{k} \right)^{k+1}
-1\right]$$ 
for all $k$. And the latter has limit $\frac{k+1}{k}$ as $x\to 0$.
So $$1 \leq \lim\limits_{x \to 0} \frac{e^{x}-1}{x} \leq \frac{k+1}{k}$$ for all $k$.
A: No: the definition
$$
e^x = \sum_{k=0}^\infty \frac{x^k}{k!}
$$
is independent of anything but convergence of series. You first prove that is converges for any real $x$, then you prove that the derivatives coincide with $e^x$ itself.
A: To find the derivative using the power series notation...
$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$
\begin{align}
\frac{d}{dx} e^x &= \sum_{k=1}^{\infty} \frac{kx^{k-1}}{k!}  \\
&= \sum_{k=0}^{\infty} \frac{(k+1)x^{k}}{(k+1)!} & \text{Transform }k-1 \mapsto k \\
&= \sum_{k=0}^{\infty} \frac{(k+1)x^{k}}{(k+1)k!} \\
&= \sum_{k=0}^{\infty} \frac{x^{k}}{k!} = e^x
\end{align}
Personally, I do not see how this is circular reasoning for the definition of the derivative. The Maclaurin series is just another representation of $e^x$, and this allows calculating $\frac{d}{dx} e^x$ by using the power rule.
