Prove that $n!e-2< \sum_{k=1}^{n}(^{n}\textrm{P}_{k}) \leq n!e-1$ Prove that $$n!e-2 < \sum_{k=1}^{n}(^{n}\textrm{P}_{k}) \leq n!e-1$$  where $^{n}\textrm{P}_k = n(n-1)\cdots(n-k+1)$ is the number of permutations of $k$ distinct objects from $n$ distinct objects and $e$ is the exponential constant (Euler's number).

Denote the expression in question by $S$
$$
\sum_{k=0}^{n-1}\frac{1}{k!}=e-\sum_{k=n}^{\infty}\frac{1}{k!}=e-\frac{1}{n!}\bigg(1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)} + \cdots \bigg)\\
\leq e-\frac{1}{n!} \bigg(1+\frac{1}{n+1} + \frac{1}{(n+1)^2} +\cdots \bigg)\\
=e-\frac{1}{n!}  \sum_{k=0}^{\infty} \bigg(\frac{1}{n+1} \bigg)^k=e-\frac{1}{n!}\bigg(1+\frac{1}{n}\bigg)
$$
Pre-multiplying this by $n!$ the upper bounds comes out easily:
$$
S<n!e-1-\frac{1}{n}
$$
