Prove: $(1-\frac{1}{d+1})^d>\frac{1}{e}$ I need to prove that $\left(1-\frac{1}{d+1}\right)^d>\frac{1}{e}$.
I guess that I have to use that $\left(1+\frac{1}{n+1}\right)^n\rightarrow e$ for $n\rightarrow\infty$ or better $<e$ or $\left(1+\frac{1}{n-1}\right)^n>e$ but I don't know exactly how it works.
Thank you for your help!

First,
$$
(1 + \frac{1}{n})^n < e,
$$
so
$$
\frac{1}{(1 + \frac{1}{n})^n} > \frac{1}{e}. 
$$
Now,
$$
(1 - \frac{1}{n + 1}) = \frac{n}{n + 1} = \frac{1}{\frac{n + 1}{n}} = \frac{1}{\left(1 + \frac{1}{n}\right)},
$$
so
$$
(1 - \frac{1}{n + 1})^n = \frac{1}{(1 + \frac{1}{n})^n} > \frac{1}{e}
$$
Q. E. D.
