$n! > (n/e)^{n}$ for all $n \geq 1$ by induction

Question

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.

Answer 1

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.

Answer 2

$$(k+1)(k/e)^k=(k+1)k^ke^{-k}\\ =(k+1)^{k+1}e^{-k-1}e^1\left(\frac k{k+1}\right)^k\\ \geq(k+1)^{k+1}e^{-k-1}e/e$$ because $e>((k+1)/k)^k\to1/e<(k/(k+1))^k$.
So the next rung of the inequality has been established.

Answer 3

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 $$ e>\frac{(k+1)^k}{k^k}=\left(1+\frac{1}{k}\right)^{\!k} $$