Dubious "proof" of $e^x$ derivative? The proof to which I am referring is amply discussed here: Derivative of exponential function proof, but I remain unconvinced by the answers that pertain to the specific proof discovered by user1346994.
It all boils down to showing that $\lim_{h\to 0}\left({\dfrac{e^{h}-1}{h}}\right) = 1$
I find it highly deceiving to replace $e$ with $(1+h)^{1/h}$ in the expression in the limit.
https://math.stackexchange.com/a/671305/309566 is the answer in which I am most interested. But I'm still puzzled by the remark 'again by continuity' and the change of variables bit. 
I suppose if my wish were modest, it would be a proof that follows the one stated but with clear justifications. Thanks and sorry if I sound confused!
 A: There have been numerous questions on MSE dealing with same problem namely the derivative of $e^{x}$. Most of the time OP (as well as the people who answer the question) assumes some definition of symbol $e^{x}$ but forget to mention the definition of $e^{x}$ explicitly.
Note that a definition of $e$ alone is not sufficient to define the symbol $e^{x}$. The linked answer assumes many things almost all of which are very difficult to establish. In particular it assumes the following:
1) Definition of $x^{h}$ for all $h$ and $x > 0$ such that it is a continuous function of $h$.
2) Existence of limit $\lim_{t \to 0}(1 + t)^{1/t}$
3) Interchange of double limits in $t, h$ assuming continuity of a complicated function of two variables $t, h$.
Without justifying the above assumptions (or at least mentioning them explicitly and the fact that they need to be justified) the answer is a classic example of intellectual dishonesty. However most elementary textbooks on calculus are guilty of the same dishonesty so no one even thinks that this is actually a problem and studying calculus with non-rigorous proofs (or even no proofs at all) is almost a tradition.
A proper proof of derivative of $e^{x}$ must begin with definition of symbol $e^{x}$ and I have provided one such approach here.
A: My attempt:
For finite $h<1$ and integer $m$,
$$f(h):=\frac{e^h-1}h=\frac{\left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^h-1}h\\
=\lim_{m\to\infty}\frac{\left(1+\dfrac1m\right)^{mh}-1}h\\
=^*
\lim_{m\to\infty}\frac{\left(1+\dfrac hm\right)^{m}-1}h\\
=\lim_{m\to\infty}1+\frac{m-1}{2m}h+\frac{(m-1)(m-2)}{3!m^2}h^2+\cdots\frac{h^m}{m^m}\\
=1+\frac h2+\frac{h^2}{3!}+\cdots$$ which is a convergent series as the terms are bounded by the powers of $h$, and the sum is bounded by 
$$\frac1{1-h}.$$
Then by squeezing
$$\lim_{h\to0}f(h)=1.$$

By a change of variable $mh\to m$. We may also start from this line, as defining $e^h$.
A: Well the first question is if you believe that $\lim_{h \rightarrow \infty}(1+h)^{1/h} = e$. Whether you mathematically believe this really comes down to how you define $e$. One definition is quite simply as this limit. You then need to prove $\frac d{dx}e^x = e^x$, which seems to be what the author is doing.
I personally prefer to define $e$ as the exponential base having self-derivative, just because this is how people have been describing $e$ to me since I was a kid. Your proof is zero steps in this case.
I think the most common way to define $e$ is by defining $\ln$ first as the area (definite integral) under $1/x$, and defining $e$ so $\ln(e) = 1$. From here, proving the limit fact is a little harder, but getting the derivative is just a matter of chasing inverses around and using the fundamental theorem of calculus.
Anyway, say you assume this limit. You can get to $h \rightarrow 0$ instead of $\infty$ easily, actually. From here it's really just composition of (continuous!) functions: if $a(x) \rightarrow a_0$ then $b(a(x)) \rightarrow b(a_0)$. What's a little trickier is establishing the limit exists at all, so yes, there is work to make this proof complete, but it's not going in the wrong direction.
A: This may not convince you either and it is not a solution to the limit problem; another way of establishing the derivative of $e^x$.Consider the inverse, $x=\ln y$, $y>0$ and differentiate both sides with respect to $x$ to get $1=\frac{1}{y}\frac{dy}{dx}$, which tells you $\frac{dy}{dx}=y$ where $y=e^x$. Then, we have $\frac{d}{dx}e^x=e^x$. We assume that we already know $\frac{d}{dx}\ln x=\frac{1}{x}$, $x>0$ here and there is an easier argument to establish that $\frac{d}{dx}\ln x=\lim\limits_{h\to0}\frac{\ln(x+h)-\ln(x)}{h}=\frac{1}{x}$, $x>0$.
