Uniform convergence of $f_n(x)=x^n(1-x)$ 
I need to show that $f_n(x)=x^n(1-x)\rightarrow0 $  is uniformly in $[0,1]$ (i.e. $\forall\epsilon>0\,\exists N\in\mathbb{N}\,\forall n>N: \|f_n-f\|<\epsilon$)

I tried to find the maximum of $f_n$, because:
$$\|f_n-f\|=\sup_{[0,1]}|f_n(x)-f(x)|=\max\{f_n(x)\}.$$
So if we investigate the maximum value of $f_n(x)$, we get:
$$f_n'(x)=0\Rightarrow x_\max=\dfrac{n}{n+1}.$$
Therefore $\|f_n\|=f_n\left(\frac{n}{n+1}\right)=\frac{n^n}{(n+1)^{n+1}}$. And here I get stuck. How can I get $\|f_n\|<\epsilon$

The $f_n$ sequence is decreasing, $[0, 1]$ is compact, constant function is continuous, so the result follows immediately from the Dini's theorem.
