proving that $(n-1)^n>n^{n-1}$ I want to prove that  $(n-1)^n>n^{n-1}$, for $n>4$, $n$ is an integer.
So I divided by $n^n$ and got:
$(1-\frac{1}{n})^{n}>\frac{1}{n}$
I know that $(1-\frac{1}{n+1})^{n+1}$>$(1-\frac{1}{n})^{n+1}=(1-\frac{1}{n})^n(\frac{n-1}{n})>\frac{1}{n}(\frac{n-1}{n})$.
How can I continue?
Thanks.
 A: take the logarithm of both sides and you will get
$\frac{\ln(n-1)}{n-1}>\frac{\ln(n)}{n}$ to prove this consider the function $f(x)=\frac{\ln(x)}{x}$
A: You could prove that $(1-\frac1n)^n$ is an increasing sequence from which your inequality easily follows. Just use the AM-GM inequality on $n$ copies of $\frac{n-1}n$ and the number $1$:
$$1-\frac1{n+1}=\frac n{n+1}=\frac{\frac{n-1}n+\ldots+\frac{n-1}n+1}{n+1}>\sqrt[n+1]{\biggl(\frac{n-1}n\biggr)^{\!n}\!\!\cdot1}=\sqrt[n+1]{\biggl(1-\frac1n\biggr)^{\!n}}$$
Therefore
$$\biggl(1-\frac1n\biggr)^{\!n}<\biggl(1-\frac1{n+1}\biggr)^{\!n+1}$$
A: A simple binomial expansion also does the trick. For $n\ge3$ it's true that
\begin{align*}(n+1)^n&=\sum_{k=0}^n\binom nk n^{n-k}<n^n+n\cdot n^{n-1}+n(n-1)\,n^{n-2}+1+\sum_{k=3}^{n-1}n(n-1)\ldots(n-k+1)\,n^{n-k}\\&\le 3n^n+1-n^{n-1}+\sum_{k=3}^{n-1}n^kn^{n-k}<3n^n+(n-3)\,n^n=n^{n+1}.\end{align*}
So your equality holds for all $n\ge 4$.
A: Solution using induction:
Base case - $n=4$, so $(4-1)^4>4^{4-1}$ which is true.
Inductive step - assume that the statement holds for some k and show that it holds for k+1, hence we want to show that $k^{k+1}>(k+1)^k$ for $k \ge 3$.
Our assumption is $(k-1)^k>k^{k-1}$ or alternatively $\displaystyle k-1>\left(1+\frac{1}{k-1}\right)^{k-1}$.
Dividing both sides by $k^k$ we get that we need to show that $\displaystyle k>\left(1+\frac{1}{k}\right)^k$ for $k \ge 3$.
It is obvious that $\displaystyle 1+\frac{1}{k}<1+\frac{1}{k-1} \rightarrow \left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{k-1}\right)^k$, so it is sufficient to show that $\displaystyle k>\left(1+\frac{1}{k-1}\right)^k$.
We can write: $$\displaystyle \left(1+\frac{1}{k-1}\right)^k=\left(1+\frac{1}{k-1}\right)\cdot{\left(1+\frac{1}{k-1}\right)^{k-1}}\underbrace{<}_{using \ assumption}\left(1+\frac{1}{k-1}\right)(k-1)=k$$
A: Another way is
to take the
$n(n-1)$
root of both sides.
It then becomes
$(n-1)^{1/(n-1)}
> n^{1/n}
$.
This is shown 
by the many proofs here
that
$n^{1/n}$
is a decreasing function
for real $n \ge e$
or integer
$n \ge 3$.
