Prove $\frac{d}{dx}(e^x)=e^x$ with Taylor series and power rule 
Prove $\dfrac{d}{dx}\left(e^x\right)=e^x$ with Taylor series and power rule.

So far I have
$$1+x+\frac{1}{2}x^2+\frac{1}{3\cdot2}x^3+\cdots+\frac{1}{\infty-1}x^\infty$$
note: Also that is what I got after I took the derivative of the Taylor series, fyi for those who are telling me to take the derivative of the Taylor series. Basically, it goes through an infinite regression.
Thence I applied the power rule upon the Taylor series of $e^x$.
I'm pretty sure I'm supposed to use the binomial theorem to get $(1+\frac{x}{n})^n$; however $n$ in this case is $\infty$, and thus I cannot rewrite the series.
Help s'il vous plaît. Merci.
 A: You know $$e^x=1+x+\frac{1}{2}x^2+\frac{1}{3\times2}x^3\dots$$
and you want to prove $\frac{d}{dx}e^x=e^x$. To compute the left hand side, differentiate the series from above.
$$\frac{d}{dx}\left(1+x+\frac{1}{2}x^2+\frac{1}{3\times2}x^3\dots\right)$$
What do you get? Do you recognise what you get?

Alternative method:
You also mentioned the formula $e^x=\lim_{n\to\infty}\left((1+\frac{x}{n})^n\right)$. You can also differentiate this with respect to $x$ by moving the derivative into the limit, and you'll get the same result.
A: $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$
the radius of convergence is $R= \infty$.
the derivative is
$$\sum_{k=1}^\infty\frac{kx^{k-1}}{k!}=$$
$$\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}=$$
$$\sum_{K=0}^\infty\frac{x^K}{K!}=e^x$$
A: Excluding the infinite term what you have there is the Taylor Series of $e^x$. You don't need to consider an infinite term, as infinite series are determined by the convergence of their partial (finite) sums. 
A: One way I like to prove it is through a differential equation:
$$f'(x)=f(x)$$
$$\frac{f'(x)}{f(x)}=1$$
$$\int\frac{f'(x)}{f(x)}dx=\int dx$$
$$\int\frac{f'(x)}{f(x)}dx=x+c_0$$
Letting $y=f(x)$ gives 
$$\int\frac{dy}{y}=x+c_0$$
$$\ln|y|=x+c_0$$
$$f(x)=e^{x+c_0}$$
Note: I actually don't know if this proof contains circular logic with the $\int\frac{dy}{y}=\ln|y|$ bit, but I hope it doesn't.
A: I suppose that $series$ and $taylor-series$ are studies after function $e^{x}$. But if you can use Taylor-representation it will be easy to differentiate series : 
$\displaystyle \frac{d (e^{x})}{dx} = \frac{d}{dx}(\sum_{k=0}^{\infty} \frac{x^{k}}{k!}) = \sum_{k=0}^{\infty}\frac{d(x^{k})}{k!dx} = \sum_{k=0}^{\infty} \frac{x^{k-1}}{(k-1)!} = \sum_{k'=0}^{\infty} \frac{x^{k'}}{k'!}$, where $k' = k-1$.
Adding : $$\lim_{\vartriangle x \to 0} \frac{e^{x+\vartriangle x} - e^{x}}{\vartriangle x} = \frac{e^{x}(e^{\vartriangle x}- 1)}{\vartriangle x}$$ and use Taylor estimation to finish limit.
A: As suggested by John Doe since
$$e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!}=\sum_{k=0}^{\infty} a_k(x)$$
and


*

*for $k=1 \implies \frac{da_1(x)}{dx}=0$

*for $k>1 \implies\frac{da_k(x)}{dx}=a_{k-1}(x)$
we have
$$\frac{de^x}{dx}=\frac{d}{dx}\left(1+\sum_{k=1}^{\infty} a_k(x)\right)=0+\frac{d}{dx}\sum_{k=1}^{\infty} a_k(x)=\sum_{k=1}^{\infty} a_{k-1}(x)=\sum_{k=0}^{\infty} a_k(x)=e^x$$
