If $n$ is a positive integer, show that $$\left(1-\frac{1}{n}\right)^n < \left(1-\frac{1}{n+1}\right)^{n+1}$$

So well if we take common denominators and simplify we get,

\begin{align}
&\frac{(n-1)^n}{n^n} < \frac{(n)^{n+1}}{(n+1)^{n+1}} \\ 
\Rightarrow &(n^2-1)^n \cdot (n+1) < (n^2)^n \cdot n \\
\end{align}

And since $(n^2)^n$ is much greater $(n^2-1)^n$ i can see that the transformed version of the inequality is true. However i couldn't come up with an actual method to prove it. I also couldn't figure out appropriate weights for the original inequality. We are supposed to use Weighted\Regular AM-GM-HM. Also maybe if we define $f(x)=\left(1-\dfrac{1}{x}\right)^x$ and show that is increasing we will be done ? However is that doable/viable approach? The main method however is supposed to be WAM-WGM-WHM. Thanks.