How to prove a polynomial can be written as Taylor-style? I know that by Taylor's theorem, a function $f$ under some assumptions, can be computed by $$f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{(n+1)}$$ If $f$ itself is a polynomial of degree $n$, then $$f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n.$$ This can be directly deduced from Taylor's theorem as I mentioned. 
However, since this is a much simpler result, can we prove it without using that theorem? And is there intuitive understanding of the above equation?
 A: Use induction on $n\in\Bbb Z_{\ge 0}$.
If $n=0$ the statement is obvious.
If the statement is true for polynomials of degree $n$ or lower and $p$ is a polynomial of degree $n+1$ then $p'$ is a polynomial of degree $n$ and
$$p'(x)=\sum_{k=0}^n \frac{p^{(k+1)}(a)(x-a)^k}{k!}$$
Therefore
$$p(x)=C+\sum_{k=0}^n \frac{p^{(k+1)}(a)(x-a)^{k+1}}{(k+1)!}$$
for some constant $C$. By taking $x=a$, we see that $C=p(a)$. Thus,
$$p(x)=\sum_{k=0}^{n+1}\frac{p^{(k)}(a)(x-a)^k}{k!}$$
A: Just observe that:
$$
D^k(x^n)\big|_{x=0}=\cases{
k! & if $n=k$,\\
0 & if $n\ne k$.\\
}
$$
It follows that if 
$$
P(x)=a_nx^n+\ldots+a_1x+a_0
$$
then
$$
P^{(k)}(0)=k!\,a_k,
\quad\hbox{that is:}\quad
a_k={P^{(k)}(0)\over k!}.
$$
An analogous reasoning can be repeated for the case 
$P(x)=a_n(x-a)^n+\ldots+a_1(x-a)+a_0$, computing derivatives at $x=a$.
A: Polynomials are already in this Taylor form.
If you want to start with a polynomial, then convert it into this Taylor form, you can do that simply by differentiating the polynomial up to $n$ times and then calculating the derivatives at $a$. If you want to prove that a polynomial is the Taylor form of a given function, then this question should help.
You might find the Taylor form more intuitive if you interpret it as a ‘standard form’ for calculus and approximations. It is easy to apply shifts and stretches to, and it can be truncated as desired to estimate a derivative at a point or to estimate the value of the original function at a point.
Does that answer your question?
A: This simply is a re-writing of the binomial formula.
First note it suffices to prove it for monomials, since differentiation is a linear operation. So let's set $f(x)=x^n$.
Second, the binomial formula yields
$$x^n=\bigl(a+(x-a)\bigr)^n=\sum_{k=1}^n\binom nk a^{n-k}(x-a)^k$$
and observe that
$$\binom nk a^{n-k}=\frac{n(n-1)\dots(n-k+1)a^{n-k}}{k!}=\frac{f^{(k)}(a)}{k!}.$$
A: For me the intuitive understanding is that the information contained in a polynomial of degree $n$ -- which can be seen as its $n+1$ coefficients, for example--, is encoded in its $n+1$ derivatives at any point. Knowing the behaviour of the polynomial at any point in full detail, you can resconstruct it entirely.
A: Let
$$f(x):=\sum_{k=0}^n c_k x^k$$
be a polynomial of degree $\leq n$, and let $a\in{\mathbb R}$ be an arbitrary point on the real axis. Then
$$g(y):=f(a+y)=\sum_{k=0}^n c_k(a+y)^k=\sum_{l=0}^n c_l' y^l\qquad(y\in{\mathbb R})$$
for certain constants $c_l'\in{\mathbb R}$, because expanding the $(a+y)^k$ with $k\leq n$ does only create powers $y^l$ with $l\leq n$. It follows that
$$f(x)=g(x-a)=\sum_{l=0}^n c_l'(x-a)^l\qquad(x\in{\mathbb R})\ .$$ For those who know higher derivatives it is easy to check that $$f^{(\ell)}(a)=l!\>c'_l\qquad(l\geq 0)\ .$$
