When $x$ is a real number and $x>1$, why is $x^x>(x+1)^{x-1}$?

By applying the weighted AM-GM to the distinct positive numbers $x+1$ and $1$ with weights $1-\tfrac1x$ and $\tfrac1x$, $$ (x+1)^{1-\frac1x}\cdot 1^{\frac1x} < \big(1-\tfrac1x\big)\cdot (x+1) + \tfrac1x\cdot 1 = x. $$ Taking $x$th powers, $$ (x+1)^{x-1} < x^x. $$

Let $f(x)=x\ln x-(x-1)\ln(x+1),\;$ so $\color{red}{f(1)=0}$

and $\displaystyle f^{\prime}(x)=x\left(\frac{1}{x}\right)+\ln x-(x-1)\cdot\frac{1}{x+1}-\ln(x+1)=\frac{2}{x+1}-(\ln(x+1)-\ln x)$.

Since $\displaystyle \ln(x+1)-\ln x<\frac{1}{x}\;\;$ (by considering the area under $y=\frac{1}{x}$ from $x$ to $x+1$),

$\displaystyle \color{red}{f^{\prime}(x)>\frac{2}{x+1}-\frac{1}{x}=\frac{x-1}{x(x+1)}>0}\;$ for $x>1$;

so for $x>1,\;\;$ $\color{red}{f(x)>0}\implies x\ln x>(x-1)\ln(x+1)\implies x^x>(x+1)^{x-1}$

Observe that if we consider $f(x)=x^x-(x+1)^{x-1}=e^{x\ln x}-e^{(x-1)\ln (x+1)}$

Then $f'(x)>0$ and $f(1)=0$.

So we can conclude that $f(x)>0$.

Hope this helps.