Show that the sequence of functions $x(1-x), x^2(1-x), x^3(1-x)...$ converges uniformly on $[0,1]$ 
Theorem: Let $E,E'$ be metric spaces, with $E'$ complete, and let $f_n: E \to E', n=1,2,3,....$ Then the sequence of functions $f_1, f_2, f_3, ...$ is uniformly convergent if and only if, for any $\epsilon >0$, there is a positive integer $N$ such that for any $x\in E$, if $n$ and $m$ are integers greater than $N$, then $d'(f_n(x),f_m(x))< \epsilon$ 

Question: Using the above theorem to show that the sequence of functions $x(1-x), x^2(1-x), x^3(1-x)...$ converges uniformly on $[0,1]$.
My proof: Given $\epsilon >0$ and let $N \in \mathbb{Z}_+$ to be determined laer, and $n,m > N$, $x\in [0,1]$. Define a function $f_n: \mathbb{R} \to \mathbb{R}$ by $f_n(x) = x^n(1-x)$.
WLOG, let $n>m$. Then,
\begin{align*}
d'(f_n(x),f_m(x))&=|x^n(1-x)-x^m(1-x)|\\
&= |x^m(1-x)(x^{n-m}-1)|\\
&=|x^m| \cdot|1-x| \cdot |1-x^{n-m}|.
\end{align*}
(1). If $x=0$, then $|x^m|=0$. Hence, $d'(f_n(x),f_m(x))=0 < \epsilon$.
(2). If $x=1$, then $|1-x|$. Hence, $d'(f_n(x),f_m(x))=0 < \epsilon$.
(3). If $x\in (0,1)$, then notice that $lim_{m\to \infty} x^m =0$. By
    definition of limit, for the given $\epsilon$, there exists an
    $N>0$ s.t. for all $m>N$, $|x^m-0|= |x^m| < \epsilon$. Also note
    that when $x\in (0,1)$, $|1-x|<1$ and $|1-x^{n-m}|<1$. Hence,
    $d'(f_n(x),f_m(x))=|x^m| \cdot|1-x| \cdot |1-x^{n-m}| < \epsilon
\cdot 1 \cdot 1 = \epsilon$.
So, we can claim that given $\epsilon >0$, there exists an $N>0$ s.t. for all $k>N$, we have $|x^k-0|= |x^k| < \epsilon$. And if $x \in [0,1]$, $n,m>N$, then $d'(f_n(x),f_m(x)) < \epsilon$. $\blacksquare$
I think my problem appears in (3) of my proof. When I say that 

For $x \in (0,1)$, we can find an $N$ such that for all $m>N$,
  $|x^m-0|= |x^m| < \epsilon$.

I have chosen a $x$ before I chose the $N$.
So, can someone write a proof by using the above theorem? I want to see how to use this theorem to prove uniform convergence.
 A: A simple argument without using any theorem on uniform convergence is as follows: given $\epsilon >0$ choose N such that $(1-\epsilon)^{N}$<$\epsilon$. Let $n>N$. Note that $|x^{n}-x^{n+1}|$ is bounded by $\epsilon$ if $x \geq \epsilon$ and bounded by $(1-\epsilon)^{n}$ if $x<\epsilon$.
A: Hint. Let $f_n(x)=x^n(1-x)$. Then
$$
f'(x)=nx^{n-1}(1-x)-x^n=x^{n-1}\big(n(1-x)-x\big)=x^{n-1}\big(n-(n+1)x\big).
$$
Hence, $f_n$ is increasing in $\big[0,\frac{n}{n+1}\big]$
and decreasing in $\big[\frac{n}{n+1},1\big]$ and thus
$$
0=f_n(0)=f_n(1) \le f_n(x)\le f_n\Big(\frac{n}{n+1}\Big)=\Big(\frac{n}{n+1}\Big)^n\Big(1-\frac{n}{n+1}\Big)<\frac{1}{n+1}.
$$
Therefore, for every $\varepsilon>0$, there exists a $N\in \mathbb N$, namely $N=\big\lfloor\frac{1}{\varepsilon}\big\rfloor$, such that
$$
n\ge N=\Big\lfloor\frac{1}{\varepsilon}\Big\rfloor\quad \longrightarrow\quad
n+1>\frac{1}{\varepsilon}\quad \longrightarrow\quad 
|\,f_n(x)|<\frac{1}{n+1}<\varepsilon,
$$
and hence $f_n\to 0$, uniformly over the interval $[0,1]$.
A: Using the AM-GM Inequality,
$$x^n(1-x)=\frac{1}{n}\,x^n(n-nx)\leq \frac{1}{n}\,\left(\frac{nx+(n-nx)}{n+1}\right)^{n+1}=\frac{n^n}{(n+1)^{n+1}}<\frac{1}{n}\,.$$
In fact, this is a plagiarism of my own answer here, as it works here as well.
