Is $ \left(\frac{1}{2}\right)^n < \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ true for $n \ge 3$?

The inequality is equivalent to $n<2(2-\frac{2}{n})^{n-1}$. For $n\geq4$ we have $2-\frac{2}{n}\geq \frac{3}{2}$. Thus it is enough to prove $n<2(\frac{3}{2})^{n-1}$, which holds for $n\geq 4$ (you can use induction to prove this if you'd like to).

We need to prove that $$(n-1)^{n-1}>\left(\frac{n}{2}\right)^n.$$ Let $f(x)=(x-1)\ln(x-1)-x\ln\frac{x}{2},$ where $x\geq3$.

Hence, $$f'(x)=\ln(x-1)+1-\ln\frac{x}{2}-1=\ln\left(2-\frac{2}{x}\right)>0.$$ Thus, $f(n)\geq f(3)=2\ln2-3\ln1.5>0$ and we are done!

If $f(x)=x(1-x)^{n-1}$ then $$f'(x)=(1-x)^{n-1}-x(n-1)(1-x)^{n-2}=(1-x)^{n-2}(1-nx)$$

So $f$ has a critical point where $x=\frac{1}{n}$. Since $f(0)=f(1)=1$ and this is the only critical point in $(0,1)$, and $f(x)>0$ in $(0,1)$, you have that $f(x)$ is maximized when $x=\frac{1}{n}.$ In particular, then:

$$f\left(\frac12\right)<f\left(\frac 1n\right)$$

This is actually true for any real $n>1$ except $n=2$ when $\frac{1}{2}=\frac{1}{n}$.

If you have a weighted coin with probability $p$ of coming up heads, then $np(1-p)^{n-1}$ is the probability than in $n$ tosses, you'd get exactly one heads. Then what the above argument says is that, to maximize the odds of getting one heads in $n$ tosses, you'd want $p=\frac{1}{n}$, which is somewhat intuitive...