Is this a valid way to prove that $\frac{d}{dx}e^x=e^x$? $$e^x= 1+x/1!+x^2/2!+x^3/3!+x^4/4!\cdots$$
$$\frac{d}{dx}e^x= \frac{d}{dx}1+\frac{d}{dx}x+\frac{d}{dx}x^2/2!+\frac{d}{dx}x^3/3!+\frac{d}{dx}x^4/4!+\cdots$$
$$\frac{d}{dx}e^x=0+1+2x/2!+3x^2/3!+4x^3/4!\cdots$$
$$\frac{d}{dx}e^x= 1+x/1!+x^2/2!+\cdots=e^x$$
Of course this proof is circular because in order to find the expansion requires knowing the derivative, but ignoring that fact, is this proof valid?

It is valid after you have shown (or defined) that $e^x$ is indeed equal to that series and that deriving the series is the same as deriving each term individually (under suitable conditions, of course). It is also not circular and works also for complex numbers.
