Do we cheat on $\int e^x$?

Question

So this question seems weird at first look, I admit that.

$f(x)=e^x$ can we written as series as follows: $f(x)=e^x = \sum \limits_{k=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+...$

Now differentiating $f(x)$ yields $e^x$ again, as we have an infinite number of summands and derivation of each summand gives the one to its left.

So far so good.

Now, basically everywhere we find this: $\int \mathrm{d}x \, e^x = e^x+C$ where C is the integration constant. Now this is, where I think we cheat a little. By integrating each summand of $f(x)$ we find, that each summand becomes the one to its right. But there's no one to the left of $1$. Do we in fact do this:

$\int \mathrm{d}x \, e^x = e^x -1 + C = e^x + \tilde{C}$ ? (with $\tilde{C}=C-1$)

Or is there another way to show, that indeed $\int \mathrm{d}x \, e^x=e^x+C$

Answer 1

Without caring about the constant: $$De^x=e^x$$ So: $$D^{-1}De^x=D^{-1}e^x$$ $$e^x=D^{-1}e^x$$ Where $D$ is the differential-operator.

Edit:
The antiderivate of $f$ is: $$\int f=\{g\,|\,g'=f\}$$ And since both $a(x)=e^x$ and $b(x)=e^x-1$ has the property that $a'=b'=\exp$, so both of them are $\in\int\exp$.

Answer 2

There is no contradiction since for Foundamental Theorem of Calculus C is an arbitrary constant thus

$$\int e^x\mathrm{d}x =(e^x-1)+C$$

and

$$\int e^x\mathrm{d}x =e^x$$

are equivalent.