Prove the inequality $0 \leq e^{-nx} - (1-x)^n \leq nx^2 e^{-nx}$ for $x \in [0,1]$ and $n \in \mathbb{N}$. This inequality appeared in a print publication:

$$0 \le e^{-nx} - (1-x)^n \le nx^2 e^{-nx} \qquad (0 \le x \le 1)$$

No conditions on $n$ were stated but it's probably safe to take $n \in \mathbb{N}$. I'm specifically interested in proving the right-hand portion of the result. Differentiation produced inconvenient expressions; however, maybe I made bad choices of functional form. Would series expansion be the preferred approach here?
 A: Recall that $e^x\ge1+x$ for all $x\in\mathbb{R}$.
We also have Bernoulli's Inequality: $(1+x)^n\ge1+nx$ for $x\ge-1$ and $n\in\mathbb{N}$.
First we have
$$
e^{-x}\ge1-x\implies e^{-nx}\ge(1-x)^n\tag{1}
$$
Next we have
$$
\begin{align}
e^x\ge1+x
&\implies e^x(1-x)\ge\left(1-x^2\right)\tag{2}\\
&\implies e^{nx}(1-x)^n\ge\left(1-x^2\right)^n\ge1-nx^2\tag{3}
\end{align}
$$
Therefore,
$$
(1-x)^n\ge e^{-nx}\left(1-nx^2\right)\tag{4}
$$
Combining $(1)$ and $(4)$, we get
$$
e^{-nx}\ge(1-x)^n\ge e^{-nx}\left(1-nx^2\right)\tag{5}
$$
which is equivalent to your inequality.
A: First, note that $1-x \le e^{-x}$ for all $x \in [0,1]$. Hence, $(1-x)^n \le e^{-nx}$, i.e. $0 \le e^{-nx}-(1-x)^n$ for all $x \in [0,1]$. 
Also, since the map $y \to ny^{n-1}$ is non-decreasing on $[0,1] \supseteq [1-x,e^{-x}]$, we have $$e^{-nx}-(1-x)^n = \int_{1-x}^{e^{-x}}ny^{n-1}dy \le \int_{1-x}^{e^{-x}}n(e^{-x})^{n-1}dy  = ne^{-(n-1)x}[e^{-x}-(1-x)] = ne^{-nx}[1-(1-x)e^x].$$
For $x \in [0,1]$, we have $e^x \ge 1+x$ and $1-x \ge 0$. Hence, $1-(1-x)e^x \le 1-(1-x)(1+x) = x^2$. 
Therefore, $e^{-nx}-(1-x)^n \le ne^{-nx}[1-(1-x)e^x] \le nx^2e^{-nx}$, which is the desired upper bound.
A: Multiply both sides by $e^{nx}$ positive for any $n,x$
Inequality becomes 
$$0\leq 1-e^{n x} (1-x)^n\leq n x^2$$
Derivative of $f(x)=1-e^{n x} (1-x)^n$ is 
$f'(x)=\dfrac{n x e^{n x} (1-x)^n}{1-x}$
as $0\leq x \leq 1$ we have $f'(x)\geq 0$ so we know that $f(x)$ is increasing and its maximum on $[0;\;1]$ is attained at $x=1$
Even at its maximum we have 
$0\leq 1-e^{1n } (1-1)^n\leq 1^2n$ that is $0\leq 1\leq n$ therefore, assuming $n\geq 1$, we have 
$$0\leq 1-e^{n x} (1-x)^n\leq n x^2, \text{ for any }x\in [0;\;1]\text{ and }n\ge 1$$
