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?

Let's define  $b^x$ as 
$$ b^x = a_0 + a_1x +a_2 \frac{x^2}{2!} + \ldots =\sum_{k=0}^{\infty} a_k\frac{x^k}{k!}$$
for $x=0$, 
$ b^0 =1 $  Thus  $a_0=1$
$$ b^{x} =  1 + a_1x +a_2 \frac{x^2}{2!} + \ldots = 1 + \sum_{k=1}^{\infty} a_k\frac{x^k}{k!} \tag 1$$
$$\frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=b^x \lim_{h \to 0} \frac{b^{h}-1}{h}=b^x \lim_{h \to 0} \frac{(1 + a_1h +a_2 \frac{h^2}{2!} + \ldots)-1}{h}$$
$$\frac{d}{dx} b^x=b^x \lim_{h \to 0}  (a_1 +a_2 \frac{h}{2!} + \ldots)=a_1b^x $$
Let's select $a_1=1$  then we define $b=e$
and then we need to find all $a_k$ values and $e$.
After selecting $a_1=1$ and $b=e$, we have :
$$\frac{d}{dx} e^x=e^x \tag 2$$
According to this defination, $a_k=1$ for $k \geq 2$
Proof:
for $a_1=1$ and $b=e$ , 
And using relation (1)
$$ e^x = 1 + x +a_2 \frac{x^2}{2!} + \ldots =1 + x+ \sum_{k=2}^{\infty} a_k\frac{x^k}{k!}$$
$$ \frac{d}{dx} e^x = 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots =$$
According to the result (2),
$$ \frac{d}{dx} e^x =e^x= 1 + x +a_2 \frac{x^2}{2!} + \ldots= 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots $$
Now We need to equal all cooeffients of $x^k$, 
we find $a_{2}=1$ and $a_{k+1}=a_{k}$  for $k \geq 2$  
If we solve that relation, we get: $a_{k}=1 $ for $k \geq 2$ 
Thus $ e^x$ can be written as power series as shown below
$$ e^x= 1 + x + \frac{x^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{x^k}{k!} $$
and to find $e$: put $x=1$, $e^1=1 + 1 + \frac{1^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{1}{k!}$

Note:
If we select $a_1=m$ and follow the same way as shown in the proof, we will get
$$ b^{x}= 1 + mx + \frac{(mx)^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{(mx)^k}{k!} =e^{mx}=(e^m)^x$$
$$ b^{x}=(e^m)^x$$
Thus
$ b=e^m$
$\ln(b)=\ln(e^m)=m$
$$\frac{d}{dx} b^x=a_1 b^x =m b^x=\ln(b) b^x $$
$$\frac{d}{dx} (b^x)=\ln(b) b^x $$
