Question: I'm trying to understand the following question

Give a new derivation of the formula
$$ \frac {\mathrm d}{\mathrm dx} e^x = e^x $$
by differentiating the power series
$$ 1 + x + \frac{x^2}2 + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots + \frac{x^k}{k!} + \cdots $$
and seeing what you get.

Do I differentiate $e^x$ and compare it to the series of $e^x$? Stuck here.

Answer: 
We have the Maclaurin-Taylor series expansion for the exponential function
$$e^x=\sum\_{k=0}^\infty \frac{x^k}{k!} = 1 + x{} + \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...$$
If we differentiate implicitly, then
$$\frac{d(e^x)}{dx} =\frac{d}{dx}\sum\_{k=0}^\infty \frac{x^k}{k!} = \frac{d}{dx} \bigg( 1 + x{} + \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+... \bigg)$$
And we know by the linearity of the differential operator, i.e. $(f(x)+g(x))' = f'(x)+g'(x)$, we get
$$\frac{d(e^x)}{dx} =\sum\_{k=0}^\infty \frac{d}{dx} \bigg( \frac{x^k}{k!} \bigg) = \sum\_{k=0}^\infty \frac{1}{k!} \frac{d}{dx} \bigg( x^k \bigg)$$
applying the constant rule for derivatives, i.e. $(cf(x))=cf'(x)$ and $(c)'=0$ ($c$ constant). Furthermore, knowing the power rule, $(x^n)' = nx^{n-1}$ and using the recursivity of the factorial, i.e. $k! = k(k-1)!$, for $k > 0$,
$$\frac{d(e^x)}{dx} = \sum\_{k=0}^\infty \frac{1}{k!} kx^{k-1} = \frac{1}{0!} 0 \cdot x^{0-1} + \sum\_{k=1}^\infty \frac{1}{k!} kx^{k-1}$$
So
$$\frac{d(e^x)}{dx}=\sum\_{k=1}^\infty \frac{1}{k(k-1)!} kx^{k-1} = \sum\_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1}$$
We will now proceed a change of index $k-1=k'$,
$$\frac{d(e^x)}{dx}=\sum\_{k'=0}^\infty \frac{1}{k'!} x^{k'}=\sum\_{k=0}^\infty \frac{x^{k}}{k!} $$
(since the terms of the sum are infinite)

Which is exactly $e^x$. Thus,
$$\frac{d}{dx} \bigg( e^x \bigg) = e^x$$
The power series of $e^x$ is given by $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum\_{k = 0}^{\infty} \frac{x^{k}}{k!}$$

Differentiating term-by-term, we see $\frac{d}{dx} \frac{x^k}{k!} = k \cdot \frac{x^{k - 1}}{k!} = \frac{x^{k - 1}}{(k - 1)!}$.

In other words, each term gets "shifted" down by one. If the sum were finite, this would mean that the last term would vanish. But since the sum is infinite, there is no "last term". Thus, we conclude $\frac{d}{dx} e^x = e^x$.