Discriminant for $x^n+bx+c$

The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern:

$$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$

corresponding to the discriminants

$$(b^2-4c, -4b^3-27c^2,-27b^4+256c^3,256b^5+3125c^4).$$

Does the pattern for the ratios extend to higher orders? (An online reference would be appreciated.)

Yes. Sketch: $b$ is a symmetric polynomial of degree $n-1$ in the roots and $c$ is a symmetric polynomial of degree $n$, whereas the entire discriminant is a symmetric polynomial of degree $n(n-1)$. It follows that the discriminant is a linear combination of $b^n$ and $c^{n-1}$, and the coefficients can be determined by setting $b = 0, c = -1$ and then $b = -1, c = 0$ and reducing to the computation of the discriminant of $x^n - 1$.

Use the relation between the disciminant of $f$ and the resultant of $f$ and $f'$. The resultant is easy to calculate since $f'$ is so simple.