All derivative of a polynomial at a point is $0$, then such polynomial is $0$ The original question I'm trying to solve:  

Let $f$ be a polynomial, suppose there exist $x_0$ such that $f^{(n)}(x_0) = 0\;\forall n \ge0$,  show $f = 0$

This question seems to be trivial via Taylor series expansion. However I'm not allowed to use that...
Really not too sure where to start with this one, any help would be appreciated, thank you.
 A: Assume that $f \not\equiv 0$. Let $a_nx^n$ be the leading term of $f$. Then $f^{(n)}(x) = n!a_n \neq 0$ for all $x$, which is a contradiction.
A: Hint: If $f(x)=\sum\limits_{k=0}^{n} a_k (x-x_0)^{n}$ then $a_0=f(x_0), a_1=f'(x_0),... a_n=\frac {f^{(n)}(x_0)} {n!}$. Can you show that any polynomial has the form $\sum\limits_{k=0}^{n} a_k (x-x_0)^{n}$? (Binomial Theorem should help in this).
Alternatively, $f(x_0)=0$ implies that $f$ has  a zero at $x_0$, $f(x_0)=f'(x_0)=0$ implies that $f$ has  a zero  of oder at least two at $x_0$, and so on. Thus $f$ has zero of any order  at $x_0$ and $(x-x_0)^{n}$ divides $f(x)$ for every $n$ (even when $n$ exceeds the degree of $f$). Hence $f\equiv 0$.
A: If $f$ is not the zero polynomial, then it has a leading term $ax^n$ for some positive integer $n$ and some non-zero real $a$.  Then $f^{(n)}(x)=a\cdot n!$ for all $x$ by repeated applications of the power rule.  But we are told that $f^{(n)}(x_0)=0$, which is a contradiction.  Therefore, $f$ can only be the zero polynomial.
