<s>[INST] Inequality understanding My textbook says that:
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n<e^n
$$
But I do not understand this. Can you please enlighten me?
Edit: How do you show that
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n
$$
 [/INST] For the first part
note that each term is 
$$\begin{align}
\left(\frac{1+k}{k}\right)^k =& \frac{(1+k)^k}{k^k}\\
=&\frac{1}{k}\frac{(1+k)^k}{k^{k-1}}\\
=&\frac{1}{k}\frac{(1+k)^k}{(1+(k-1))^{k-1}}
\end{align}$$ 
so every denominator cancels nicely with the previous numerator:
$$ \begin{align}\frac{1}{n!}\left(\frac{(1+1)^1}{(1+0)^0}\frac{(1+2)^2}{(1+1)^1}\frac{(1+3)^3}{(1+2)^2}\cdots\frac{(1+n)^n}{n^{n-1}}\right)\\
=\frac{1}{n!}(1+n)^n\end{align}$$
For the second part
note that $f(k) = \left(1+\frac{1}{k}\right)^k < e$ for every $k>0$ since $f(k)$ is a monotonically increasing function and $\lim_{k\to\infty}f(k) = e$, so that the product of any $n$ of them is less than $ee\cdots e = e^n$.
</s>