TITLE: A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$
QUESTION [6 upvotes]: Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.

REPLY [4 votes]: 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.