Evaluating Nested Limits In my attempt to prove $\frac d{dx}\left(e^x\right)=e^x$, I arrived at the following step:
$$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{e^h-1}{h}$$
At this point, I substituted in 
$$e=\lim_{n\to\infty}{\left(1+\frac 1n\right)^n}$$
to arrive at
$$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{\lim_{n\to\infty}{\left(1+\frac 1n\right)^{nh}}-1}{h}$$
Then, I let $h=\frac 1n$ to get
$$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{(1+h)^{h\left(\frac 1h\right)}-1}{h}$$
at which point a lot of things simplify and the limit is clearly equal to one. My question concerns the $h=\frac 1n$ step. While I see why this might be true, I don't really have a good conceptual or mathematical grasp of why this works, or whether this is even valid at all. Can the nested limit just be removed that way and $n$ replaced, or is there something else at work here?
 A: It is not legitimate to let $h=1/n$ since $h$ and $n$ are independent parameters.  We can still proceed along any of the following ways forward.

METHODOLOGY $1$:
In THIS ANSWER, I showed that $\left(1+\frac xn\right)^n$ converges uniformly to $e^x$ on any bounded interval of the real line.  
Therefore, the function $\frac{d}{dx}\left(1+\frac xn\right)^n=\left(1+\frac xn\right)^{n-1}$ converges uniformly to $e^x$ on any bounded interval of the real line.
Hence, $\frac{de^x}{dx}=e^x$ for all $x\in\mathbb{R}$.

METHODOLOGY $2$:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities 
$$1+x\le e^x\le \frac{1}{1-x}$$
for $x<1$.  Therefore, we can write for $0<h<1$
$$1\le \frac{e^h-1}h\le \frac{1}{1-h}$$
whence application of the squeeze theorem yields the coveted limit
$$\lim_{h\to 0^+} \frac{e^h-1}h=1$$
A parallel development for $h<0$ results in the same conclusion.  And we are done.

METHODOLOGY $3$:
In THIS ANSWER, I discuss the Moore-Osgood Theorem, which permits the interchange of limits if one of the two limits converges pointwise and the other uniformly.
It can be shown that $\lim_{h\to 0}\frac{\left(1+\frac1n\right)^{nh}-1}{h}=n\log\left(1+\frac1n\right)$ and the convergence is uniform for $n\in \mathbb{N}$.
Hence, the Moore-Osgood Theorem guarantees that 
$$\begin{align}
\lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac1n\right)^{nh}-1}{h}&=\lim_{n\to \infty}\lim_{h\to 0}\frac{\left(1+\frac1n\right)^{nh}-1}{h}\\\\
&=\lim_{n\to \infty} n\log\left(1+\frac1n\right)\\\\
&=1
\end{align}$$
as expected again!
