Proving $k^k>(k+1)^{k-1}$ For $k>2,$ Show $k^k<(k!)^2$
Using induction, I boiled it down to proving $k^k>(k+1)^{k-1}$
Im stuck here. I tried using binomial inequality that i.e. $$k^k=((k+1)-1)^k>(k+1)^k-k(k+1)^{k-1}=(k+1)^{k-1}$$
But the fact that it has a negative number may not be the right argument, so, how to proceed?
 A: HINT
We have that
$$k^k>(k+1)^{k-1}\iff k\log k>(k-1)\log(k+1)\iff\frac{\log k}{k-1}>\frac{\log(k+1)}{k}$$
then consider $f(x)=\frac{\log x}x\implies f'(x)=\frac{1-\log x}{x^2}<0$ as $x>e$.
A: Note that $\frac{(k+1)!)^2}{k!^2}=(k+1)^2$ and that$$\frac{(k+1)^{k+1}}{k^k}=(k+1)\left(1+\frac1k\right)^k.$$Now, use the fact that$$(k+1)\left(1+\frac1k\right)^k<(k+1)\times e$$in order to prove that $\frac{(k+1)^{k+1}}{k^k}<\frac{(k+1)!)^2}{k!^2}$.
A: Without using calculus:


*

*You can write $$(k!)^2=\prod_{n=1}^{k}n(k-n+1)$$ that is, $(k!)^2=(1\cdot k)\times (2\cdot (k-1))\times \ldots \times (k\cdot 1)$.

*Show that $n(k-n+1)\geq k$. This is the same as showing that, when $1\leq n\leq k$, $$n^2-n(k+1)+k=(n-1)(n-k)\leq 0$$ which is easily seen to be true since $n-k$ is always negative or $0$, and $n-1$ is always positive or $0$.
Notice that $(n-1)(n-k)<0$ (strict inequality) if $n\neq1$ and $n\neq k$. Because of this, we get our strict inequality precisely when there is atleast one $n$ such that $1<n<k$ (which fails when $k=2$).
Thus, when $k>2$, $$(k!)^2=\prod_{n=1}^{k}n(k-n+1)> \prod_{n=1}^{k}k=k^k$$
