Find $n,m \geq 1$ s.t. $p(x) = 1 - (1 - x^n)^m$ satisfying $p(\frac{c}{2}) < \epsilon$ and $p(c) > 1 - \epsilon$ Having trouble showing this lemma and its corollary within Stone-Weierstrass. My instructor said this was easy to show, so he did not show it in class. But I cannot seem to do it.

Lemma: Let $\epsilon >0$ and $0 < c < 1$. There exists $n,m \geq 1$ s.t. $p(x) = 1 - (1 - x^n)^m$ satisfying $p(\frac{c}{2}) < \epsilon$ and $p(c) > 1 - \epsilon$.


Corollary: Let $p(x)$ be given as in the above lemma. Then $0 \leq p(x) < \epsilon$ where $x \in [0,\frac{c}{2}]$ and $p(x) \geq 1 - \epsilon$ where $x \in [c,1]$.

I feel like the lemma is more of a statement, and not a lemma. It seems like he is just giving us a function, $p(x)$, and telling us how it is defined. I looked back through my notes, and asked him, and he said that is all we need. So I've kind of ignored that, but I'm just not getting anywhere with the corollary. If I plug in $0$, I get $0$, the lower bound. So I'm assuming by plugging in $c/2$, I have to show the function has its maximum between $0$ and $\epsilon$ at $c/2$.
Similar for the next one, I have to show that at $c$ the function is equal to $1 - \epsilon$ and then I know it's maximum is $1$ from the definition in the lemma. I'm just not sure how to do it, or if my thoughts are on the right track.
 A: Corollary:
For the corollary notice that for all $x \in [0,1]$
$$
p'(x)=mnx^{n-1}(1-x^n)^{m-1} \geq 0.
$$
So $p$ is non-decreasing in $[0,1]$. Notice that $p(0)=0<p(\frac{c}{2})<\epsilon$. Hence $0 \leq p(x)<\epsilon$ for all $x \in [0,\frac{c}{2}]$. And notice that $p(c) \geq 1- \epsilon$. Hence $p(x) \geq 1- \epsilon$ for all $x \in [c,1]$.
Without using differantiation:
Suppose $x<y \in [0,1]$. Then $x^n<y^n$ and $-x^n>-y^n$. Hence $1-x^n>1-y^n$ and $(1-x^n)^m>(1-y^n)^m$. So $-(1-x^n)^m<-(1-y^n)^m$ and $1-(1-x^n)^m<1-(1-y^n)^m$. This implies that $p(x)<p(y)$ and $p$ is increasing.
Lemma:
This is my attempt on proving the Lemma where I used Bernoulli's inequality which is a standard result and with an easy proof. Hope it helps and if there is something wrong with the proof please let me know.
Edit: Bernoulli's inequality states that
$$
(1+x)^r \geq 1+rx
$$
for any integer $r \geq 0$ and for every real number $x > -1$. For a proof of Bernoulli's inequality take a look at Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$. It is a simple application of induction.
Notice that we can suppose $\epsilon<1$. Now, since $0<c<1$, we know that $\frac{1}{c} > 1$ and we can choose $k$ the smallest positive integer such that $k>\frac{1}{c}$. Then we have
$$
k-1 \leq \frac{1}{c} \implies k \leq \frac{1}{c}+1<\frac{2}{c}.
$$
Also notice that $-\frac{c^n}{2^n}> -1$ for any $n \in \mathbb{N}$. Taking $x=-\frac{c^n}{2^n}$ and $r=k^n$ we can apply Bernoulli's inequality
$$
\left(1+\left(-\frac{c^n}{2^n}\right) \right)^{k^n} \geq 1-k^n\frac{c^n}{2^n}=1-\left(k\frac{c}{2} \right)^n
$$
Then follows from $k<\frac{2}{c}$ that $\delta_0=(k\frac{c}{2}) < 1$ and
\begin{equation}
\left(1-\frac{c^n}{2^n} \right)^{k^n} \geq 1-{\delta_0}^n.
\end{equation}
Now let's use Bernoulli's inequality again using that $c^n > -1$ for any $n \in \mathbb{N}$. This time take $x=c^n$ and $r=k^n$ and apply the inequality
$$
(1-c^n)^{k^n} = \frac{1}{(kc)^n}(1-c^n)^{k^n}k^nc^n \leq \frac{1}{(kc)^n}(1-c^n)^{k^n}(1+k^nc^n)
$$
$$
(1-c^n)^{k^n} \leq \frac{1}{(kc)^n}(1-c^n)^{k^n}(1+c^n)^{k^n}=\frac{1}{(kc)^n}(1-c^{2n})^{k^n}<\frac{1}{(kc)^n}.
$$
First equality holds because we multiplied and divided by $k^nc^n$ and first equality holds because $k^nc^n<1+k^nc^n$. Then we just used Bernoulli's inequality on $1+k^nc^n$ and used the fact that $0<1-c^{2n}<1$.
Now using that $k>\frac{1}{c}$ we have $\delta_1=\frac{1}{kc}<1$. Then
\begin{equation}
(1-c^n)^{k^n}<{\delta_1}^n.
\end{equation}
Notice that ${\delta_0}^n \to 0$ and ${\delta_1}^n \to 0$ as $n \to \infty$ (because $0<\delta_0,\delta_1<1$). Hence, using the definition of limit, we can choose $n_0$ such that ${\delta_0}^{n_0}<\epsilon$ and ${\delta_1}^{n_0}<\epsilon$. From what we already proved we have
$$
\left(1-\frac{c^{n_0}}{2^{n_0}} \right)^{k^{n_0}} \geq 1-{\delta_0}^n>1-\epsilon
$$
$$
(1-c^{n_0})^{k^{n_0}} < {\delta_1}^n<\epsilon.
$$
By choosing $n=n_0$ and $m=k^{n_0}$ we have
$$
p \left(\frac{c}{2} \right)=1-\left(1-\frac{c^{n_0}}{2^{n_0}}\right)^{k^{n_0}}<1-(1-\epsilon)=\epsilon
$$
$$
p(c)=1-(1-c^{n_0})^{k^{n_0}}>1-\epsilon.
$$
