Given $a,b_0,\ldots,b_n$, there exists a polynomial of degree $\le n$ s.t. the derivatives $f^{(i)}(a)=b_i$ Just exploring some maths problems from a book until I came across this question. 
Let $a, b_0, . . . , b_n ∈ R$. Show that there exists a polynomial $f(x)$ of degree at most n such that
$$f(a) = b_0, f'
(a) = b_1, f''(a) = b_2,\ldots , f^{(n)}
(a) = b_n$$
I am not sure how to approach this problem, can someone give me a guide of how to do this problem? 

Consider the following polynomial:
$$ p(x) = \sum\limits_{i=0}^n \frac{b_i}{i!}(x-a)^i $$
And observe that:
$p^{(k)}(a) = \left(\sum\limits_{i=0}^{k-1} 0\right) + \sum\limits_{i=k}^n\frac{b_i}{(i-k)!}(x-a)^{i-k} = b_k $
