Let $n >2$ be an integer, and let $0<s \le \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ be real.
Is there a way to find explicitly the unique solution $0<a<\frac{1}{n}$ to the equation $a(1-a)^{n-1} = s$?
This is the smallest root of the polynomial $p(a)=a(1-a)^{n-1}$.
The condition $s \le \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ comes from the fact that $$\max_{a \in (0,1)}a(1-a)^{n-1}=\frac{1}{n}(1-\frac{1}{n})^{n-1},$$
so if $s > \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ there is no solution in $(0,1)$. Moreover, since the function $f(a)=a(1-a)^{n-1}$ is increasing in $(0,\frac{1}{n}]$, we see that if $0=f(0)<s<f(\frac{1}{n})= \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ then there exist a unique solution $a \in (0,\frac{1}{n}]$ as mentioned above.
Comment: This question is related to this problem.
Of course, there is no general formula for the roots of polynomials of order 5 and higher, but this is a very specific polynomial, and I am asking only for the smallest root.
If an explicit formula is out of reach, what estimates can be done? Is there a known method for estimating the smallest root of a polynomial?
I am interested in estimates from above and from below.
Edit: It seems any formula will be extremely messy, even for $n=3$.