calculus - 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?

Answer

Suppose you have an exponential function, like $f(x)=2^x$.

The derivative is

$$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{2^{x+h}-2^x}{h} = \lim_{h\to0}\left(2^x\cdot\frac{2^h-1}{h} \right).$$
So far, just algebra. Now watch this:
$$= 2^x\lim_{h\to0}\frac{2^h-1}{h}.$$
This can be done because $2^x$ is "constant" and "constant" means "not depending on $h$".

But this is equal to $(2^x\cdot\text{constant})$. But in this case "constant" means "not depending on $x$". "Constant" always means "not depending on something", but "something" varies with the context.

What's the "constant"? In the case above, it's not hard to show that the constant is somewhere between $1/2$ and $1$. It we'd started with $4^x$ instead, then it would be fairly easy to show that the "constant" would be more than $1$. For a base somewhere between $2$ and $4$, the "constant" is $1$. That base is $e$.

If you want to talk about how it is knonw that $2$ is too small and $4$ is too big, to be the base of the "natural" exponential function (i.e., the one for which the "constant" is $1$), I can post further on this.

Later edit: OK, how do we know that $2$ is too small and $4$ is too big, to serve in the role of the base of the "natural" exponential function? Look at the graph of $y=2^x$. It gets steeper as you go from left to right. As $x$ goes from $0$ to $1$, $y$ goes from $1$ to $2$. So the average slope between $x=1$ and $x=2$ is

$$\dfrac{\text{rise}}{\text{run}} = \frac{2-1}{1-0} = 1.$$
Since it gets steeper going from left to right, the slope at the left end of this interval, i.e. at $x=0$, must be less than that. Thus we have $\dfrac{d}{dx}2^x = (2^x\cdot\text{constant})$ and the "constant" is less than $1$. (Thinking about the interval from $x=-1$ to $x=0$ in the same way shows that the "constant" is more than $1/2$.)

Now look at $y=4^x$, and use the interval from $x=-1/2$ to $x=0$, and do the same thing, and you see that when you write $\dfrac{d}{dx}4^x = (4^x\cdot\text{constant})$, then that "constant" is more than $1$.

This should suggest that $4$ is too big, and $2$ is too small.

You can show that $3$ is too big by using the interval from $x=-1/6$ to $x=0$ and doing the same thing. But the arithmetic is messy.