Differentiating the Taylor expansion of $e^x$

Question

It is well known that a) $\frac{d}{dx}\exp x = \exp x$ and b) $\exp x = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + ...$. Therefore, it should be possible to differentiate this Taylor series representation of $\exp$ and get back the same function. Doing this:

$$\frac{d}{dx}\exp x = \frac{d}{dx}\sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = \sum\limits_{n=0}^\infty \frac{nx^{n-1}}{n!} = \sum\limits_{n=0}^\infty \frac{x^{n-1}}{(n-1)!}$$

This is fairly unsurprising, however this has a problem - consider the differentiated series at $n=0$. This term simplifies (trivially) to "$\frac{1}{x(-1)!}$". $(-1)!$ is now involved with this. There seems be no obvious reason to change the starting index to 1 either.

Answer 1

For the case where $n=0$, you cannot cancel $n$ on the top and bottom because $0!=1$ has no factor of $0$. I suggest making the $n=0$ case separate.

$$\exp{x}=1+\sum_{n=1}^{\infty}\frac{x^n}{n!}$$

Then after deriving, just shift your index.

Answer 2

Your mistake is, basically, that you differentiated all factors $\frac{x^n}{n!}$ and got $\frac{x^{n-1}}{(n-1)!}$, when in fact, this is only true for $n>0$. For $n=0$, this is not true:

$$0=\left(\frac{x^0}{0!}\right)' \neq \frac{x^{-1}}{(-1)!}$$