Polynomial nth derivative I was wondering how is this done?
Let $a,b_0,...,b_n \in \mathbb{R}$. 
Show that there exists a polynomial $f(x)$ of degree at most $n$ such that $f(a) = b_0, f'(a) = b_1, f''(a) = b_2, ..., f^{(n)}(a) = b_n$
I can intuitively see that for $x^n$ all the derivatives of it would give a real number when x is a real number, however how do I show what the question is asking for?

Recall the structure of the Taylor series centered at $x=a$:
$$f(a)+f'(a)(x-a)+\cdot\cdot\cdot+\dfrac{f^n(a)}{n!}(x-a)^n+\cdot\cdot\cdot$$
While of course the function is just a polynomial of $n$ degrees and not a Taylor series, the convenience with this series is that $f^n(a)$ can be found easily.
Note: For the polynomial of $n$ degrees, we can have a generic polynomial
$f(x)=a_n(x-a)^n+\cdot\cdot\cdot+a_1(x-a)+a_0 \quad a_i\in \mathbb{R},\quad\!\! i\in\{1,2,\cdot\cdot\cdot,n\}$
$f(a)=a_0=b_0$
$f'(a)=a_1=b_1$
$\quad\cdot$
$\quad\cdot$
$\quad\cdot$
$f^n(a)=a_n=b_n$
This reveals something special about the statement "show that there exists a polynomial of degree at most $n$ such that conditions are true."
