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.

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: $$\displaystyle \left(1+\frac{1}{k-1}\right)^k=\left(1+\frac{1}{k-1}\right)\cdot{\left(1+\frac{1}{k-1}\right)^{k-1}}\underbrace{<}_{using \ assumption}\left(1+\frac{1}{k-1}\right)(k-1)=k$$
