Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$ 
Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$.

My try: $f'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}(n-(n+1)x)=0\Rightarrow x=\frac{n}{n+1} $, hence $\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}$. I know $(1-\frac{1}{n+1})^n\to e^{-1}$. But I don't know how to prove $(1-\frac{1}{n+1})^n\lt \frac{1}{e}$.
 A: Hint: Prove that $g(x)=x^x$ is increasing, then with $n<n+1$
$$(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}=\dfrac{n^n}{(n+1)^{n+1}}<1$$
Edit: You want to prove
$$\left(1+\dfrac{1}{n}\right)^{n+1}>e$$
by mean value theorem for $\ln x$
$$\ln(n+1)-\ln(n)=\dfrac{1}{\xi}>\dfrac{1}{n+1}$$
where $n<\xi<n+1$.
A: \begin{equation}
 \max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{n}{n+1})^n\cdot\frac{n}{n+1}
 =
 \frac{n}{(n+1)^{n+1}}
 \tag{1}
\end{equation}
Now
\begin{equation}
 \frac{n}{(n+1)^{n+1}} - \frac{1}{ne}
 =
 \frac{1}{n}( \frac{n^2}{(n+1)^{n+1}} - \frac{1}{e})
 =
 \frac{1}{n}( \frac{en^2 - (n+1)^{n+1}}{e(n+1)^{n+1}})
 \tag{2}
\end{equation}
Sign depends on $en^2 - (n+1)^{n+1}$. It is easy to see that for $n=1$,  $en^2 - (n+1)^{n+1} < 0$. We can say that for $n \geq 2$,
\begin{equation}
\begin{split}
  en^2 - (n+1)^{n+1} &<  e(n+1)^2 - (n+1)^{n+1} \\ &= (n+1)^2(e - (n+1)^{n-1}) <0
\end{split}
\end{equation}
So for all $n \geq 1$, we have that $en^2 - (n+1)^{n+1} < 0$. Going back to $(2)$, we get $\frac{n}{(n+1)^{n+1}} - \frac{1}{ne} < 0$, which using $(1)$, gives 
\begin{equation}
 \max_{x\in(0,1)} f(x)
 <
 \frac{1}{ne}
\end{equation} 
