$\star$ Show that the probability of no successes in $n$ Bernoulli trials, each with probability $p = 1/n$, is approximately $1/e$. Show that the probability of exactly one success is also $1/e$

The probability of no successes of $n$ trials is:

$$ b(0;n,p) = \binom{n}{k} p^k q^{n-k} = \binom{n}{0} q^n = (1-p)^n = \Big(1 - \frac{1}{n}\Big)^n = \frac{1}{\Big(1-\frac{1}{n}\Big)^{-n}} $$

The probability of one success of $n$ trials is:

$$ b(1;n,p) = \binom{n}{1} \frac{1}{n} \bigg(1 - \frac{1}{n}\bigg)^{n-1} = \frac{1}{\Big(1-\frac{1}{n}\Big)^{-(n-1)}} $$

We know that:

$$ e = \lim_{n \to \infty}\bigg(1 + \frac{1}{n}\bigg)^n = \lim_{n \to \infty}\bigg(1 - \frac{1}{n}\bigg)^{-n} $$

We can just substitute in the examples above and get approximately $1/e$.