Prove if a power series is zero on an interval then all coefficients are zero

A power series as the same radius of convergence as the series of derivatives. In particular, we can take the derivative term by terms for $|x|<c$. We have $P^{(k)}(0)=0$, and translating this in term of $\sum_k a_kx^k$, this means that $a_k=0$.

Assume otherwise and let $n$ be moinimal with $b_n\ne 0$. Then $Q(x)=\sum_{k=0}^\infty b_{n+k}x^k$ converges for $|x|<1$ and we have $P(x)=x^nQ(x)$ and hence also $Q(x)=0$ for $0<|x|<c$. by continuity, also $Q(0)=0$, i.e. $b_n=0$ - contradiction.

Let $f(x) = \sum_{k=0}^\infty c_k x^k$ be a power series with radius of convergence $R \gt 0$ and $f(x)=0$ for all $|x| \lt S$ where $0 \lt S \lt R$, then $c_k=0$ for all $k$.

Consider the $n$-th derivative at the point $x=0$. We get

$$f^{(n)}(0) = \sum_{k=n}^\infty \frac{k!}{(k-n)!}c_k0^{k-n} = n!c_{n}$$

since all summands cancel except of the first. Keep in mind that we defined for power series that $0^0=1$ (but only for power series!).

Since we know that $f(x)=0$ on $(-S,S)$, we know that its derivative must also be zero on the same interval. Therefore it follows that $c_n=0$ for all $n$.

But since each coefficient is zero, the whole power series is the zero function and in fact zero on all of $\Bbb R$.