Proving $f : (-1, 1) \rightarrow \mathbb{R}$ has a property I'd like to answer the two questions below. I believe my solution to the first one is right, but it'd help if someone could please help verify this, as I am working through this book independently through self-study. I don't know how to do the second one, which is a continuation of the first problem. I'm pretty sure both questions utilize a Theorem stated in my book, so I've provided that as well.

Suppose $f : (-1, 1) \rightarrow \mathbb{R}$ has $n$ derivatives and its $n^{\text{th}}$, suppose its $n^{\text{th}}$ derivative $f^{(n)} : (-1, 1) \rightarrow \mathbb{R}$ is bounded. Finally, suppose we have
$$f(0) = f'(0) = f''(0) = \cdots f^{(n - 1)}(0) = 0.$$
Prove there exists a positive number $M$ such that $$|f(x)| \leq M|x|^{n}.$$

Second question:

Suppose $f : (-1, 1) \rightarrow \mathbb{R}$ has $n$ derivatives and there is a positive number $M$ such that $$|f(x)| \leq M|x|^{n}.$$ Prove $$f(0) = f'(0) =  \cdots f^{(n - 1)}(0) = 0.$$

Also, here's the theorem I was referring to:

Theorem: Let $I$ be an open interval and $n$ be a natural number and suppose that the function $f : I \rightarrow \mathbb{R}$ has $n$ derivatives.  Suppose also at the point $x_{0}$ in $I$, $$f^{(k)}(x_{0}) = 0$$ for $0 \leq k \leq n - 1$. Then, for each point $x \neq x_{0}$ in $I$, there is a point $z$ strictly between $x$ and $x_{0}$ at which $$f(x) = \frac{f^{(n)}(z)}{n!}(x - x_{0})^{n}.$$


Here's my attempt at problem $1$:
By the Theorem above, for each $x \in (-1, 1)$, $x \neq 0$, there's a point $z$ between $x$ and $0$ such that $$f(x) = \frac{f^{(n)}(z)}{n!}x^{n}.$$
Using the fact that $f^{(n)}$ is bounded, we know there exists a bound $N$ such that $|f^{(n)}(x)| \leq N$ for all $x$ in $(-1, 1)$.  Therefore,
$$|f(x)| = \left|\frac{f^{(n)}(z)}{n!}x^{n}\right| \leq |\frac{N}{n!}x^{n}| = (N/n!)|x|^{n} $$
So setting $M = N/n!$ suffices.
Is this correct? How do I do the next one?

For the second question: Suppose the result fails. Then $f^{(k)}(0)\ne 0$ for some $k\in \{0,1,\dots,n-1\}.$ Let $k_0$ be the smallest such $k.$ We then have $f^{(k_0)}(0)\ne 0$ and $f^{(k)}(0)= 0$ for $k<k_0.$
By the theorem you cited, for each $x\in (-1,1)\setminus\{0\}$ there exists $z_x$ beween $0$ and $x$ such that
$$f(x) = \frac{f^{(k_0)}(z_x)}{k_0!}x^{k_0}.$$
It follows that for such $x.$
$$M|x|^n \ge \frac{|f^{(k_0)}(z_x)|}{k_0!}|x|^{k_0}.$$
Therefore
$$Mk_0!|x|^{n-k_0} \ge |f^{(k_0)}(z_x)|.$$
Now $n>k_0.$ Thus as $x\to 0,$ the left side $\to 0.$ Because $x\to 0$ forces $z_x\to 0,$ the right side $\to |f^{(k_0)}(0)|.$ The reason I can say the last is that $f^{(k_0)}$ is differentiable on $(-1,1),$ hence is continuous there.
We have thus shown $f^{(k_0)}(0)=0,$ contradiction. Thus there is no such $k_0,$ and this gives the desired conclusion.
