$n^n<(n-1)^{n+1}$ for integer $n\ge5$
I'd write $m=n-1$. Then the statement reduces to $$\left(1+\frac1m\right)^m<\frac{m^2}{m+1}.$$ It's well-known that $(1+1/m)^m$ increases to $e$, but more naively, $$\left(1+\frac1m\right)^m=1+1+\frac1{m^2}{m\choose 2} +\frac1{m^3}{m\choose 3}+\cdots<1+1+\frac12+\frac16+\cdots<3.$$ But $$\frac{m^2}{m+1}>\frac{m^2-1}{m+1}=m-1>3$$ if $n>4$.
Here is a proof without induction.
$(1-1/n)^n$ is strictly increasing for $n > 1 $ and approaches $1/e$ as $n \to \infty$ whereas $1/(n-1)$ is strictly decreasing and approaches $0$ as $n \to \infty$. Hence there is an $N_0$ such that for all $n > N_0$,
$$ \Big(1 - \frac{1}{n}\Big)^n > \frac{1}{n-1} $$
By little computation, we find that smallest $n$ for which this inequality holds is $n = 5$. Since LHS is increasing and RHS is decreasing, it implies that the inequality will never be violated for $n \ge 5$.
The claim $x^x<(x-1)^{x+1}$ is equivalent to $$ x\ln x<(x+1)\ln(x-1). $$ Consider $f(x)=(x+1)\ln(x-1)-x\ln x$ and prove that
- $f'(x)=\ln\left(1-\frac1x\right)+\frac2{x-1}\to 0$ as $x\to +\infty$,
- $f''(x)=-\frac{x+1}{x(x-1)^2}<0$ for $x>1$.
Then $f'(x)>0$ and $f$ increases. From $f(5)>0$ it follows that $f(x)>0$ for all $x\ge 5$.