Prove: $(1-\frac{1}{d+1})^d>\frac{1}{e}$ I need to prove that $\left(1-\frac{1}{d+1}\right)^d&gt;\frac{1}{e}$.

I guess that I have to use that $\left(1+\frac{1}{n+1}\right)^n\rightarrow e$ for $n\rightarrow\infty$ or better $&lt;e$ or $\left(1+\frac{1}{n-1}\right)^n&gt;e$ but I don't know exactly how it works.

Thank you for your help!
Approach 1

Using Bernoulli's Inequality, which is strict for $d\ge2$,
$$
\begin{align}
\frac{\left(1-\frac1{d\vphantom{+1}}\right)^{d-1}}{\left(1-\frac1{d+1}\right)^d}
&amp;=\frac{\left(\frac{d-1}{d\vphantom{+1}}\right)^{d-1}}{\left(\frac{d}{d+1}\right)^d}\\
&amp;=\frac{d}{d-1}\left(\frac{d^2-1}{d^2}\right)^d\\[9pt]
&amp;\gt\frac{d}{d-1}\left(1-\frac1d\right)\\[15pt]
&amp;=1
\end{align}
$$
Thus, $\left(1-\frac1{d+1}\right)^d$ is strictly decreasing and its limit is $\frac1e$. Therefore, we have
$$
\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e}
$$



Approach 2

If $x\ge-n$, Bernoulli's Inequality says that for $x\ne0$ and $n\ge2$,
$$
1+x\lt\left(1+\frac x2\right)^2\le\left(1+\frac xn\right)^n
$$
Therefore, taking the limit as $n\to\infty$, we get that for $x\ne0$,
$$
1+x\lt e^x
$$
Thus,
$$
1+\frac1d\lt e^{1/d}
$$
Raise both sides to the $-d$ power
$$
\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e}
$$
Well if you know l'hospitals rule you can show that $$(1-\frac{1}{x})^{x}$$ is equal to $e^{-1}$ as $x \to \infty$
and use this fact to prove your equation. If you don't, e can be defined as $$lim_{x-&gt;\infty}(1+\frac{1}{x})^{x}$$ and the function is less than that of e
Use this facts plus  $$lim_{x-&gt;\infty}(1+\frac{a}{x})^{x}=e^{a}$$ (try to prove this yourself as exercise) 
Substitute the $x$ for $d+1$ and the $a$ for -1 all this guides you to the answer