How to use the power series of $e^x$ to find the derivative of $e^x$? I am currently reading Roger Penrose's The Road to Reality and in the book, the author poses various problems with three different levels of difficultly easy, hard and really hard, according to the author this is easy. The problem I am looking at is as follows:

Using the power series of $e^x$ show that $de^x = e^x \, dx$

I have no idea as to how to tackle this problem.
If someone could provide some key points to solving the problem that would be great. Please do not provide the full steps, just key ideas or things to note.
Thanks!
EDIT:
I believe I understand this now because when you take the derivative of a power series you can do it term by term. The power series for $e^x$ is:
$$e^x = \sum_{i=0}^\infty \frac{x^n}{n!}$$
But more expanded it looks like this:
$$e^x =  1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} +\ldots $$
If the derivative of each term then I get: $0 + 1 + x + \frac{x^2}{2} + \ldots$ So in essence, I'm coming back to the original series. Therefore, the derivatives are the same.
 A: The power series representation of $e^x$ is
$$
e^x = \sum_{i=0}^{\infty} \frac{x^n}{n!}
$$
When you take the derivative of a power series you can do it term by term:
$$\begin{align}
\frac{d}{dx}e^x &= \frac{d}{dx}\sum_{i=0}^{\infty} \frac{x^n}{n!} \\
&= \sum_{i=0}^{\infty} \frac{d}{dx}\frac{x^n}{n!} \\
&= \sum_{i=1}^{\infty}\frac{x^{n-1}}{(n-1)!}
\end{align}
$$
This is the same as $e^x$ (think about it).
To add a bit more detail. Remember that there is a rule that says that 
$$
\frac{d}{dx} x^n = nx^{n-1}.
$$
You for example have that 
$$
\frac{d}{dx} x^2 = 2x \quad\quad\frac{d}{dx} x^3 = 3x^2.
$$
So in the above we have used that 
$$\frac{d}{dx}\frac{x^n}{n!} =\frac{1}{n!} \frac{d}{dx} x^n = \frac{1}{n!}nx^{n-1} = \frac{n}{n\cdot(n-1)!}x^{n-1} = \frac{x^{n-1}}{(n-1)!}.
$$
A: Just write it out,
$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\;,$$
and differentiate it term by term. What’s the derivative of $\dfrac{x^n}{n!}$?
A: If you are having trouble to visualize this series with sum notation, just write down term by term and derive them. Doing this you will get the idea.
$\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$
$\displaystyle \frac{d}{dx}e^x=\frac{d}{dx}\big(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\big)=$
$\displaystyle =\frac{d}{dx}\big(1\big)+\frac{d}{dx}\big(\frac{x}{1!}\big)+\frac{d}{dx}\big(\frac{x^2}{2!}\big)+\frac{d}{dx}\big(\frac{x^3}{3!}\big)+\frac{d}{dx}\big(\frac{x^4}{4!}\big)+\ldots=$
$\displaystyle =0+\frac{1}{1!}\frac{d}{dx}(x)+\frac{1}{2!}\frac{d}{dx}(x^2)+\frac{1}{3!}\frac{d}{dx}(x^3)+\frac{1}{4!}\frac{d}{dx}(x^4)+\ldots=$
$\displaystyle =\frac{1}{1!}+\frac{1}{2!}2x+\frac{1}{3!}3x^2+\frac{1}{4!}4x^3+\ldots=$
$\displaystyle =\sum_{n=1}^\infty \frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$
I hope this clarify all the process done.
A: We denote $\displaystyle f_n(x)=\frac{x^n}{n!}$, so we have $\displaystyle e^x=\sum_{n=0}^{\infty}f_n(x).$
Let $[a,b]$ an arbitrary interval.
To justify the term by term differntiation of the series on $[a,b]$ we verify this points:


*

*the series $\displaystyle \sum_{n=0}^{\infty}f_n(x_0)$ at some point $x_0\in[a,b]$ and 

*the series of derivatives $\displaystyle \sum_{n=0}^{\infty}f'_n(x)$ converges uniformly on $[a,b]$ to, say, $g$,


in this case, the series $\displaystyle \sum_{n=0}^{\infty}f_n(x)$ converges at every $x\in [a,b]$ and 
$$\displaystyle \left(\sum_{n=0}^{\infty}f_n(x)\right)'=g(x).$$
In our question, the two points are straightforward,  for instance, we verify the second point:
we have $$|f'_n(x)|=\frac{|x|^{n-1}}{(n-1)!}\leq \frac{\max(|a|,|b|)^{n-1}}{(n-1)!}=c_n,$$
so the series $\displaystyle\sum_nf'_n(x)$ converges uniformly on $[a,b]$ by Weierstrass M-test since the series $\displaystyle\sum_n c_n$ is convergent.
A: If you want an `elementary' proof which doesn't have to use the fact that you can differentiate Taylor series term by term within the radius of convergence, you can argue as follows.
By definition
$$ \frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} =  \lim_{h \to 0} \frac{e^{x}(e^{h} - 1)}{h} = \lim_{h \to 0} \frac{e^{x}\left( \sum_{n=0}^\infty \frac{h^n}{n!}- 1\right)}{h}$$
substituting in the series definition of $e^h$.
We can then cancel out the first term with the $-1$ and multiply out the factor of $h$ to obtain
$$ \frac{d}{dx} e^x =  e^x \lim_{h \to 0} \sum_{n=1}^\infty \frac{h^{n-1}}{n!} = e^x \lim_{h \to 0} (1 + h.y) $$
where $y := \sum_{n=0}^\infty \frac{h^n}{(n+2)!}$. However, by the triangle inequality we have
$$|y| \leq \sum_{n=0}^\infty \frac{|h|^n}{(n+2)!}\leq \sum_{n=0}^\infty \frac{|h|^n}{n!} = e^{|h|}$$
and so $\lim_{h \to 0} h.y =0$, since $h \to 0$ and $|y|$ is bounded.
It follows that
$$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} (1 + h.y) =e^x.$$
