Proof of "Taylor Series" We know that , according to the Taylor Series :

$$f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots \tag{1}$$

And then Maclaurin series as $a=0$.
But how to prove it?
I have googled it and searched for it in my books but I really have not come across any where where it gives the proof of $(1)$.
Please help ... Just links to the full proof is also appreciated..
Thanks!!
 A: No, we don't.
Taylor's theorem states that 
$$f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\ldots+\frac{f^{(n)}(a)}{n!}(x-a)^n+ R_n(x)$$
where quite a few things can be said about $R_n(x)$, for example
$$ \lim_{x\to a}\frac{R_n(x)}{(x-a)^n}=0$$
or
$$ \forall x\exists \xi\colon (a<\xi<x\lor x<\xi<a)\land R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$$
But we do not have in general that $\lim_{n\to\infty}R_n(x)=0$, hence not necessarily 
$$f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots $$
A: The way you state this is to imprecise to give a proof. The correct statement is that $$f(x) =\sum_{k=0}^N \frac{f^{(k)}(0)}{k!}x^k +r_N(x,\ldots)$$
Without the last term, the so called remainder (for which various different expressions are known, see this page, the statement is not correct. The proof of the statement with remainder (usually by induction) is rather easy using the fundamental theorem of calculus or the mean value theorem.
Edit (the representation of $f$ by an infinite sequence of this kind is only true for so called real analytical functions)
A: Take a look at chapters 7 and 11 (Specially section 11.9) of the well-known book:
Calculus, Vol. 1: One-Variable Calculus, with an Introduction to Linear Algebra
by Tom. M. Apostol.
There is a rigorous study of Taylor polynomials and series in those chapters.
