Differentiating the exponent power series We know that
$$
e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}
$$
We know that the series is uniformly convergent everywhere, and therefore we can differentiate term by term, i.e
$$
\left(\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}\right)' = \sum\limits_{n=0}^{\infty}\frac{nx^{n-1}}{n!}
$$
But the above is equal to $e^x$. What will happen if I do this $n$ times?
$$
(e^x)^{(n)} = \sum\limits_{n=0}^{\infty}1
$$
What is wrong here?
 A: There are a couple of minor inaccuracies in mathematical claims:


*

*The exponential series
$$
\sum_{k=0}^{\infty} \frac{x^{k}}{k!}
\tag{1}
$$
isn't uniformly convergent on the entire real line (or complex plane), but is uniformly convergent on compact sets (i.e., "uniformly convergent on compacta").

*For termwise differentiation, the important fact is that the derived series converges uniformly on compacta. Here, the derived series is the original series, so you're perfectly correct that the exponential series can be differentiated term by term.
Now for your actual question: The issue is that you haven't "simplified" after differentiating:
\begin{align*}
\frac{d}{dx}\sum_{k=0}^{\infty} \frac{x^{k}}{k!}
  &= \sum_{k=0}^{\infty} \frac{d}{dx}\, \frac{x^{k}}{k!}
  && \text{termwise differentiation} \\
  &= \sum_{k=0}^{\infty} \frac{kx^{k-1}}{k!} && \\
  &= \sum_{k=1}^{\infty} \frac{kx^{k-1}}{k!} && \text{the $k = 0$ term vanishes} \\
  &= \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k-1)!} && \text{cancellation} \\
  &= \sum_{k=0}^{\infty} \frac{x^{k}}{k!} && \text{change of index.}
\end{align*}
That is, the derived exponential series is precisely the exponential series. Differentiating finitely many times, say $n$ times, has exactly the same effect as differentiating once, as expected.
A: You can't differentiate it $n$ times, because $n$ is the index of the sum. This would mean you would differentiate the first term once, the second term twice, the third therm trice, the fourth term four times, etcetera. But you can't do a similiar thing to the left handed side. So it doesn't make sense. 
A: I think the source of your confusion is the wonderfully compact $\sum_{n=0}^\infty$ notation. I always unpack it when I have to manipulate it. Then 
$$ 
\sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6} + \cdots
$$
Written that way, there is no $n$ on the right hand side. When you differentiate term by term (as many times as you wish) the series reproduces itself (every time).
