What's wrong with this "proof" that $\frac {de^x}{dx} \neq e^x$?
$$e^x = \sum _{n=0}^\infty \frac{1}{n!}x^n $$ Therefore $$\frac{d e ^x}{d x}= \sum _{n=0}^\infty \frac{n}{n!}x^{n-1}= \frac{1}{(-1)!}x^{-1}+\sum _{n=0}^\infty \frac{1}{n!}x^n\neq e^x$$
The mistake must be somewhere in the fact that I pull the derivative operator "through" the sum operator. That is possible if the sum is finite, but given that it's infinite it must be invalid, but why?
Notice that $$e^x = \sum _{n=0}^\infty \frac{1}{n!}x^n = \frac{1}{0!}x^0 + \sum _{n=1}^\infty \frac{1}{n!}x^n = 1 + \sum _{n=1}^\infty \frac{1}{n!}x^n, $$
since $0! = x^0 = 1$. Now, we can derive:
$$\frac{d e^x}{dx} = \frac{d}{dx}\left(1 + \sum _{n=1}^\infty \frac{1}{n!}x^n\right) = \frac{d}{dx}1 + \frac{d}{dx}\sum _{n=1}^\infty \frac{1}{n!}x^n = \\ 0 + \sum _{n=1}^\infty \frac{d}{dx}\frac{1}{n!}x^n = \sum _{n=1}^\infty \frac{n}{n!}x^{n-1} = \sum _{n=1}^\infty \frac{1}{(n-1)!}x^{n-1}.$$
At this point, if you pose $m = n-1$, then you get that:
$$\frac{d e^x}{dx} = \sum _{n=1}^\infty \frac{1}{(n-1)!}x^{n-1} = \sum _{m=0}^\infty \frac{1}{m!}x^{m},$$
which is what you are looking for.