Help find the derivative of $e^{2^x}$ using the definition of the derivative Let $f(x) = e^{2^x}$, where $e$ is the exponential function.
So the $f'(x)$ is:

$\begin{align}f'(x) &=& \lim_{h \to 0} \frac{e^{2^{x+h}}-e^{2^x}}{h}\\
                      &=& \lim_{h \to 0} \frac{e^{2^{x}2^h}-e^{2^x}}{h}
\end{align}$

I don't know how to finish from here.
I believe the next step is to remove the $h$ in the denominator which means I need to factor out the $h$ in the numerator but I am not sure how to do it as that $h$ is an exponent of an exponent.
I know I can use the chain rule to arrive at $e^{2^x}2^x\ln 2$ but I would like to try it this way for practice.
 A: here is the proof:
prerequisite:
$\begin{align}e=\lim_{x\to 0}(1+x)^{\frac{1}{x}}\end{align}$  from wiki
now let's begin:
$\begin{align}f'(x) &=& \lim_{h \to 0} \frac{e^{2^{x+h}}-e^{2^x}}{h}\\
&=& e^{2^x}\lim_{h \to 0} \frac{e^{2^{x}(2^h-1)}-1}{h}\\
\end{align}$
now let $2^h-1=y$ then $y=\frac{\ln{(y+1)}}{\ln2}$, and we know when $h\to0$, then $y\to0$. then we have:
$\begin{align}f'(x) 
&=& e^{2^x}\lim_{y \to 0} \frac{e^{(2^{x}y)}-1}{\frac{\ln(y+1)}{\ln2}}\\
&=& e^{2^x}\ln2\lim_{y \to 0} \frac{e^{(2^{x}y)}-1}{\ln(y+1)}\\
\end{align}$
Now we expand $e$ by prerequisite:
$\begin{align}f'(x) 
&=&e^{2^x}\ln2\lim_{y \to 0} \frac{(1+y)^{\frac{1}{y}2^{x}y}-1}{\ln(y+1)}\\
\end{align}$
and further let $z=2^x\ln(y+1)$, or equivalently $y+1=e^{\frac{z}{2^x}}$, when $y\to0$,$z\to0$. we get:
$\begin{align}f'(x) 
&=&e^{2^x}\ln2\lim_{z \to 0} \frac{(e^{\frac{z}{2^x}})^{2^{x}}-1}{\frac{z}{2^x}}\\
&=&e^{2^x}\ln2*2^x\lim_{z \to 0} \frac{e^{z}-1}{z}\\
&=&e^{2^x}\ln2*2^x\lim_{z \to 0} \frac{((1+z)^{\frac{1}{z}})^{z}-1}{z}\\
&=&e^{2^x}\ln2*2^x
\end{align}$
