A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$

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 $n \ge 2$.

Hint: Just expand it using the Binomial Theorem, and use the obvious fact that for $i < n$, $i< n$.

In this answer, and in this answer for $x=1$, it is shown, using the binomial theorem, that for $x\ge0$, $$ \begin{align} \left(1+\frac xn\right)^n &=\sum_{k=0}^n\binom{n}{k}\frac{x^k}{n^k}\\ &=\sum_{k=0}^n\left(\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dots\frac{n-k+1}{n}\right)\frac{x^k}{k!}\\ &\le\sum_{k=0}^n\frac{x^k}{k!} \end{align} $$ where the inequality is strict for $x\gt0$ and $n\gt1$.

Setting $x=1$ gives your result.