$n! > (n/e)^{n}$ for all $n \geq 1$ by induction I am stuck on this problem at the last part of the $p = n+1$ step.
For $n = 1$ we get $1! > 1/e$, which checks out. So assuming $n=k$ for $k \geq 1$ is true, I want to show that $ (k+1)! > ((k+1)/e)^{k+1} $. So to show this I begin by expanding the expression leading to $(k+1)k! > (k+1)(k/e)^{k}$. Now im stuck, so I want to find an expression on the right side to use the fact that $ e > (1+(1/n))^{n}$ - How does this help me reach the necessary step? Insight on this would be greatly appreciated.
 A: $$(k+1)(k/e)^k=(k+1)k^ke^{-k}\\
=(k+1)^{k+1}e^{-k-1}e^1\left(\frac k{k+1}\right)^k\\
\geq(k+1)^{k+1}e^{-k-1}e/e$$
because $e>((k+1)/k)^k\to1/e<(k/(k+1))^k$.
So the next rung of the inequality has been established.
A: Let's show that
$$
(k+1)\left(\frac{k}{e}\right)^{\!k}>\left(\frac{k+1}{e}\right)^{\!k+1}
$$
which is equivalent to
$$
\frac{k^k}{e^k}>\frac{(k+1)^k}{e^{k+1}}
$$
that is
$$
e>\frac{(k+1)^k}{k^k}=\left(1+\frac{1}{k}\right)^{\!k}
$$
A: You are assuming
$$n!>(n/e)^n$$
and you want to show 
$$(n+1)!>\left ( \frac{n+1}{e} \right )^{n+1}.$$
We want to manipulate both sides to look more like the statement that we have assumed. The left side is $(n+1)n!$. The right side is $\frac{n+1}{e} \left ( \frac{n}{e} \right )^n \left ( \frac{n+1}{n} \right )^n$. By the inductive hypothesis you can replace $\left ( \frac{n}{e} \right )^n$ by $n!$, since that only makes the right side larger. This reduces the problem to showing 
$$n+1>\frac{n+1}{e} \left ( \frac{n+1}{n} \right )^n.$$
Divide both sides by $n+1$, multiply both sides by $e$, and long divide in the parentheses; the result is
$$\left ( 1 + \frac{1}{n} \right )^n<e$$
which is a pretty classic problem.
