Prove by Induction that $n! > \left(\frac{n}{e}\right)^n$ for $n>1$ I'm studying a combinatorics text (Cameron) in preparation for taking discrete math this upcoming semester. I've taken a fair amount of math in college, through Calculus II, but this is my first 500+ level math course.  
Here is the problem that I'm stuck on. It appears after an introduction to Induction proofs. 
Prove by Induction that: 
$$n! > \left(\frac{n}{e}\right)^n$$
for $n>1$. 
The book gives a hint: may use the fact that $\left(1 + \frac{1}{n}\right)^n < e$ for all n. 
I have the base step: 
$$P(1): 1! > \left(\frac{1}{e}\right)^1$$
$$1 > 0.37$$
For the inductive step, assume: 
$$P(k): k! > \left(\frac{k}{e}\right)^k$$
and prove: 
$$(k + 1)! > \left(\frac{k+1}{e}\right)^{k + 1}$$
 A: Hint.  Start be re-expressing the right hand side for $k+1$ to get
$$\Bigl(\frac{k+1}{e}\Bigr)^{k+1}
  =\frac{k+1}{e}\Bigl(\frac{k+1}{k}\Bigr)^k\Bigl(\frac ke\Bigr)^k
  =\frac{k+1}{e}\Bigl(1+\frac{1}{k}\Bigr)^k\Bigl(\frac ke\Bigr)^k\ .$$
Now use the hint you were given, and the inductive assumption.
A: Notice that
$$(k+1)!=(k+1)k!>(k+1)\left(\frac ke\right)^k>\frac{k+1}e\left(\frac{k+1}e\right)^k=\left(\frac{k+1}e\right)^{k+1}$$
The last inequality step uses the fact that
$$k^k>\frac{(k+1)^k}e$$
or,
$$e>\frac{(k+1)^k}{k^k}=\left(1+\frac1k\right)^k$$
A: Hint:  $$\dfrac{\left(\dfrac{k+1}{e}\right)^{k+1}}{\left(\dfrac{k}{e}\right)^{k}}=\dfrac{k+1}{e}\left(\dfrac{k+1}{k}\right)^k=\dfrac{k+1}{e}\left(1+\dfrac{1}{k}\right)^k$$ while 
$$\dfrac{(k+1)!}{k!}=k+1$$
A: For the proof by induction, assume that 
$$k!>\left(\frac{k}{e}\right)^k$$
and show that this implies
$$(k+1)!>\left(\frac{k+1}{e}\right)^{k+1}$$
you can express this inequality as:
$$(k+1)k!>\left(\frac{k+1}{e}\right)\left(\frac{k+1}{e}\right)^k$$
divide by $(k+1)$:
$$k!>\left(\frac{1}{e}\right)\left(\frac{k+1}{e}\right)^k$$
When this is combined with the initial assumption, it's apparent that proving the following inequality to be true will complete the proof by induction:
$$\left(\frac{k}{e}\right)^k>\left(\frac{1}{e}\right)\left(\frac{k+1}{e}\right)^k$$
refactor:
$$\frac{k^k}{e^k}>\frac{(k+1)^k}{e^{k+1}}$$
rearrange:
$$e>\left(\frac{k+1}{k}\right)^k=\left(1+\frac{1}{k}\right)^k$$
This last inequality is already given by your hint, so the proof is complete.
