Never forget the basics , many faults??

True or False?

Recall that $e^x = 1 + x +\dfrac{x^2}{2!} +\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots.$ Differentiating with respect to $x,$ we obtain $\dfrac{d\,(e^x)}{dx} = 0+ 1 +x + \dfrac{x^2}{2!} +\dfrac{x^3}{3!}+\cdots =e^x.$ And integrating, we obtain $\displaystyle {\int e^x = x + \dfrac{x^2}{1\cdot2!} +\dfrac{x^3}{3\cdot 3!} +\dfrac{x^4}{4\cdot 4!} +\cdots +C},$ where $C$ is the constant of integration. Then it must be true that $0+ 1 +x + \dfrac{x^2}{2!} +\dfrac{x^3}{3!}+\cdots = x + \dfrac{x^2}{1\cdot2!} +\dfrac{x^3}{3\cdot 3!} +\dfrac{x^4}{4\cdot 4!} +\cdots +C$ because the derivative of $e^x$ is the same as that of the integration of $e^x$ .