A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$ Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.
 A: 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 $n \ge 2$.
A: Hint: Just expand it using the Binomial Theorem, and use the obvious fact that for $i < n$, $i< n$.
A: 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.
A: The binomial expansion tells you the following $$(1+\frac{1}{n})^n= 1^n + n*1^{n-1}*\frac{1}{n}+\binom{n}{2}*1^{n-2}*(\frac{1}{n})^2 + \cdots +(\frac{1}{n})^n $$ $$= 1+ 1+ \frac{1}{2!}(1-{1\over n}) + \cdots + \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1 - \frac{n-1}{n})$$ $$\lt 1+1+\frac{1}{2!}+\cdots+\frac{1}{n!} $$
I recommend you to do the calculation to see how it does this, it really helps in the future.
