I want to compute the discriminant of $x^{n+1}+x$.
It's easy to see the roots of this polynomial are $0$ and $e^{(2k+1)\pi i/n},k=0,1,\dots,n-1$. But it's still quite difficult, if we compute all $\sigma_k(c_1,\dots,c_n),k=1,2,\dots,n$ and use Newton's formula to compute all $s_k$.
Is there any easy way to do this?
On
We can expand the resultant of $f$ and $f'$ along the first line (following the notations here) to get : $$\operatorname{res}(f,f') = (-1)^{n}\det \left(\begin{array}{c|c} I_n & I_n \\\hline I_n & (n+1) I_n\end{array}\right)$$ The formula for $2\times 2$ block determinant then gives : $$\operatorname{res}(f,f') = (-n)^n$$
Therefore, we have : $$\operatorname{dis}(f) = (-1)^{\frac{n(n+1)}{2}}\operatorname{res}(f,f') =(-1)^{\frac{n(n-1)}{2}}n^n$$
Use the properties of discriminants and resultants: for any monic polynomial $f$ of degree $d$, $$\operatorname{disc}f=(-1)^{d(d-1)/2}\operatorname{res}(f,f')=(-1)^{d(d-1)/2}\prod_{x:f(x)=0}f'(x).$$ For our $f(x)=x^{n+1}+x$, the roots of $f(x)=0$ are $x=0$ (then $f'(x)=1$) and the $n$ roots of $x^n=-1$ (then $f'(x)=-n$). Thus we get $\operatorname{res}(f,f')=(-n)^n$ and $\color{blue}{\operatorname{disc}f=(-1)^{n(n-1)/2}n^n}$.