Where does this inequality come from for $e^{-2}$? One of the proofs in my textbook has the following step involving an inequality for $e^{-2}$:
$$
\left(1 - \frac1{d+1} \right)^{2d} \ge \frac1{e^2}
$$
Where does this come from?
 A: As usual everything follows from $e^x \geq 1+x$.
For $x \in \mathbb{R}$ and $n \in \mathbb{N}$ such that $1 + \frac{x}{n} > 0$ this inequality leads to
$$
e^x = e^{\frac{x}{n} \cdot n} \geq \left(1 + \frac{x}{n} \right)^n
$$
and taking reciprocals
$$
e^{-x} \leq \left(1 + \frac{x}{n} \right)^{-n} = \left(1 - \frac{x}{n + x}\right)^n.
$$
Take $x=1$ to get
$$ \frac{1}{e} \leq \left(1 - \frac{1}{n+1}\right)^n
$$
and therefore
$$ \frac{1}{e^2} \leq \left(1 - \frac{1}{n+1}\right)^{2n}.
$$
A: You have to know that $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$$
Then you can manipulate your expression and find 
$$\left(1-\frac{1}{n+1}\right)^{2n} = \underbrace{\left[\left(1+\frac{1}{-(n+1)}\right)^{-(n+1)}\right]^{-2}}_{\to\frac{1}{e^2}\;\text{as $n$ increases}}\underbrace{\left(1+\frac{1}{-(n+1)}\right)^{-2}}_{\to1\;\text{as $n$ increases}}$$
The limit to $n\to\infty$ of the first underbraced term is $e^{-2}$ and the limit of the second right-hand term is 1. Since the left-hand side term is monotically decreasing, you know that
$$\left(1-\frac{1}{n+1}\right)^{2n} \geq \frac{1}{e^2}$$
