Differentiation of every order and Taylor series Let $f(x)$ be a function defined in $(-1,1)$ with derivatives of all orders at zero equal to zero;
that is: $f'(0)=0 , f''(0)=0 , f'''(0)=0 ...$
If there exists $c>0$ such that:  $Sup|f^{(n)}(x)| <= n!c^n$ with $x \in (-1,1)$ and $n \in \mathbb{N}$ where $f^{(n)}(x)$ is the $n$th-derivative of $f(x)$ and $Sup|f^{(n)}(x)|$ is the supremum of $n$th-derivative of $f(x)$ in absolute value.
Prove that $f(x)=0$ for any $x \in (-1,1)$.
 A: We have via Maclaurin's series for $f(x)$ the following relation $$f(x) = f(0) + xf'(0) + \cdots + \frac{f^{(n - 1)}(0)}{(n - 1)!}x^{n - 1} + \frac{f^{(n)}(\xi_{n})}{n!}x^{n}$$ and by the hypotheses of the question we get $$f(x) = \frac{f^{(n)}(\xi_{n})}{n!}x^{n}\tag{1}$$ and this leads to $$|f(x)| \leq |cx|^{n}\tag{2}$$ Let $k$ be a number with $0 < k < 1$. Then we can see that $$|f(x)| < k^{n}$$ if $|x| < k/c$. Since $k^{n} \to 0$ as $n \to \infty$ it follows that $f(x) = 0$ for all $x$ with $|x| < k/c$. If $k/c \geq 1$ then we are done.
Suppose that $b = k/c < 1$ then we need slightly more work. Clearly we can see that all derivatives of $f$ are zero when $|x| < b$. And since each of these derivatives is continuous it follows that all these derivatives are zero if $|x|\leq b$. Consider $g(x) = f(x + b)$, so that all derivatives of $g$ at $0$ are $0$ and derivatives of $g$ satisfy the same inequality as $f$. Hence we can repeat the same logic and it will be found that $g(x) = 0$ for all $x$ with $0 \leq x \leq k/c = b$. So $f(x) = 0$ for all $x$ with $0 \leq x \leq 2b$. Continuing in this fashion we can extend further and show that $f(x) = 0$ for all $x$ with $0 \leq x \leq nb$ for any positive integer $n$ provided $f$ is defined in that range.
Similar exercise can be done for negative values of $x$ and our job is done. Note that the interval $(-1, 1)$ was arbitrary and result would be valid for any interval containing the point $0$.

Update: Based on comment from OP regarding the value of $f(0)$ I have added some remarks. The answer presented above assumes that $f(0) = 0$. This is not given as a hypothesis in the question explicitly. However if $f(0) \neq 0$ then it is not possible anyway to get the conclusion $f(x) = 0$ for all $x \in (-1, 1)$. Let's then find out what happens when $f(0) \neq 0$. In that case consider the function $F(x) = f(x) - f(0)$. Clearly $F(0) = 0$ and all derivatives of $F$ are $0$ at $0$. Hence the argument of my answer above applies to it and $F(x) = 0$ for all $x$ wherever it is defined. This means that $f(x) = f(0)$ so that $f(x)$ is a constant for all values of $x$ wherever it is defined.
