Proof of $\frac{d}{dx}e^x = e^x$ I'm working through the proof of $\frac{d}{dx}e^x = e^x$, and trying to understand it, but my mind has gotten stuck at the last step.

Starting with the definition of a derivative, we can formulate it like so:

$$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$

After some algebra, we arrive at:

$$\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}$$

As $h\to0$, the expression approaches $\frac{0}{0}$, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot $\frac{e^x - 1}{x}$ and see the function's behavior at $0$," which ends up approaching $1$, which can substitute the limit to give the result of the derivative:

$$\frac{d}{dx} e^x = e^x \cdot 1 = e^x$$

I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of $1$ is obtained?
Everything in the "proof" will depend on your definition of the function $e^x$. I will choose the definition 
$$
e^h \overset{\text{def}}{=} \sum_{k=0}^\infty \frac{h^k}{k!}.
$$
Using this, one sees that 
$$
\frac{e^h - 1}{h} = \frac{\sum_{k=0}^\infty \frac{h^k}{k!} - 1}{h} = \sum_{k=1}^\infty \frac{h^{k-1}}{k!} = 1 + h \sum_{k=0}^\infty \frac{h^k}{(k+2)!}.
$$
If you have studied convergence tests, you know that the last series on the right converges for all $h \in \mathbb R$, hence taking the limit when $h \to 0$, the RHS goes to $1$ because the series will converge to something and the $h$ factored out will make the product go to zero.

Another way to do this would be to show that this series is an analytic function and is its own Taylor expansion around zero (this is not hard to do using convergence tests), so to differentiate it you can go term by term and readily see that its derivative is itself.

A third approach, which will sound a little stupid and meaningless but is nonetheless funny, is choosing another definition for $e^x$ : consider the differential equation 
$$
f'(x) = f(x), \quad f(0) = 1.
$$
Using differential equation theory it is really not hard at all to show that the solutions to the equation (without the initial condition) is a one-dimensional vector space and there exists an unique element of this vector space which satisfies the initial condition $f(0) = 1$, because the solutions are of the form $Cg(x)$ for some solution $g(x)$. Let $exp(x)$ be defined as a solution to this differential equation satisfying the initial condition. Then clearly $exp'(x) = exp(x)$. Then you can easily see that $exp'(x)$ is a differentiable function, so by induction $exp(x)$ is an infinitely differentiable function with Taylor expansion
$$
\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}.
$$
I mentioned it to show the importance of which definition one decides to choose ; it can change the whole structure of an argument.

Hope that helps,
l'Hôpital's rule applies to indeterminate form of the limit you have in
$$
\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}
$$
So you get:
$$
\lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h-1)}{\frac{dh}{dh}}=\lim_{h \to 0}\frac{e^h}{1}=1
$$