summation - A binomial inequality with factorial fractions: $left(1+frac{1}{n}right)^n

Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in \mathbb{N}$.

Answer

We have by the binomial identity that
$$\begin{align*} \left(1 + \frac 1n \right)^n &= \sum_{k=0}^n \binom nk \frac 1{n^k}\\ &= \sum_{k=0}^n \frac{n!}{(n-k)! n^k} \cdot \frac 1{k!}\\ &= \sum_{k=0}^n \frac{n \cdot (n-1) \cdots (n-k+1)}{n \cdot n \cdots n} \cdot \frac 1{k!}\\ &\text{now the first factor is $<1$ for $k\ge 2$}\\ &< \sum_{k=0}^n \frac 1{k!} \end{align*}$$
for $n \ge 2$.