Discriminant of a trinomial 
Let $b,c \in \mathbb{Z} $ and let $n \in \mathbb{N} $, $n \ge 2. $ Let $f(x) = x^{n} -bx+c$. Prove that $$\hbox{disc} (f(x)) = n^{n }c^{ n-1}-(n-1)^{n-1 }b^{n }.$$

Here $\hbox{disc} (f(x)) = \prod_{i} f'(\alpha_{i} )$ where $\alpha_{1}, \dots, \alpha_{n}$ are the roots of $f(x)$.
After some calculations I obtained $\hbox{disc} (f(x)) = \frac{\prod_{i} \alpha_{i}(n-1)b \ - \ nc }{\prod_{i} \alpha_{i}} $, but I'm afraid this is the wrong way.

Here is a brute force approach:
$f'(x) = nx^{n-1} - b$, and we want to compute $\prod_i f'(\alpha_i)$. We do this by looking for the minimal polynomial with roots $\alpha_i^{n-1}$.
Note that
$$\begin{array}%
x^n - bx+c = 0 &\Leftrightarrow x(x^{n-1} - b) = -c \\
&\Leftrightarrow x^{n-1} (x^{n-1} - b)^{n-1} = (-1)^{n-1}c^{n-1}
\end{array}$$
let $y_i = \alpha_i^{n-1}$, and $z_i = f'(\alpha_i) = ny_i - b$. We have found the minimal polynomial for $y_i$:
$$y(y-b)^{n-1} = (-1)^{n-1}c^{n-1}$$
and we want the product $\prod_i z_i$. Consider the change of variable $z = ny-b$, i.e. $y = \frac{z+b}{n}$. Substitute into the above equation, we get
$$\frac{z+b}{n} \left(\frac{z - (n-1)b}{n}\right)^{n-1} = (-1)^{n-1}c^{n-1} \\
\Leftrightarrow (z+b)(z-(n-1)b)^{n-1} - (-1)^{n-1} n^n c^{n-1} = 0$$
The constant term of this polynomial in $z$ is $(-1)^n \prod_i z_i$, therefore
$$\prod_i z_i = (-1)^n \left((-1)^{n-1}b^n(n-1)^{n-1} - (-1)^{n-1} n^n c^{n-1}\right) = n^n c^{n-1} - (n-1)^{n-1}b^{n}$$
