Showing that $p^n(1-p) \leq \frac{1}{en}$ I am reading a paper and found the following Lemma without a proof.
Let $X_1, \ldots, X_{n+1}$ be independent Bernoulli random variables, where $\Pr[X_i = 1] = p$. Let $E$ be the event that the first $n$ variables are all $1$, but the $X_{n+1}$ is $0$.
Then $\Pr[E] \leq \frac{1}{en}$.
I understand that $\Pr[E] = p^n(1-p)$. How is it that $p^n(1-p) \leq \frac{1}{en}$?

Using the AM-GM for first inequality below, we have
$$
p^n(1-p)=n^n\left[(p/n)^n(1-p)\right]\leq n^n\left(\frac{1}{n+1}\right)^{n+1}=\frac{1}n\left(1-\frac{1}{n+1}\right)^{n+1}<\frac{1}{ne}.
$$
The second inequality uses the fact that $a_n=(1-1/n)^n$ increases in $n$ and converges to $1/e$ as $n\to\infty$.
