Discriminant of a trinomial
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 $$\prod_i z_i = (-1)^n \left((-1)^{n-1}b^n(n-1)^{n-1} - (-1)^{n-1} n^n c^{n-1}\right) = n^n c^{n-1} - (n-1)^{n-1}b^{n}$$
Two solutions. The first is based on resultants. As I wasn't 100% confident in my handling of the leading coefficients, I calculated the discriminant also using the definition. I am keeping the first (worse) solution for "educational purposes".
A promising way is to use the description of the discriminant as a resultant of $f(x)$ and $f'(x)$. The resultant is (when non-zero) also a generator of the ideal $I=(f(x),f'(x))\cap\mathbb{Z}$ and can be calculated with steps like the Euclidean algorithm, where we need to take care to keep everything as integers (necessitating appropriate scaling).
Here we first calculate $$ r_1(x)=nf(x)-xf'(x)= -b(n-1)x +cn $$ and are "lucky" to get a linear remainder. Proceeding we see that $$ g(x)=(cn)^{n-1}-[b(n-1)x]^{n-1} $$ is a multiple of $r_1(x)$ by a polynomial in $\mathbb{Z}[x]$. We can then eliminate $x$ by calculating the combination $$ r_2(x)=(b(n-1))^{n-1}f'(x)+ng(x)=n^nc^{n-1}-(n-1)^{n-1}b^n. $$
Here's a more convincing calculation. Let us write $$ \begin{aligned} f(x)&=(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_n),\\ f'(x)&=n(x-\beta_1)(x-\beta_2)\cdots (x-\beta_{n-1}). \end{aligned} $$ Here the numbers $\beta_j, j=1,2,\ldots,n-1,$ are the zeros of $f'(x)$, i.e. we have $\beta_j=\zeta^j K$, where $K=\root{n-1}\of {b/n}$ and $\zeta=e^{2\pi i/(n-1)}$ is an appropriate root of unity.
Now $$ \begin{aligned} \operatorname{disc}(f)&=\prod_i f'(\alpha_i)=n^n\prod_{i,j}(\alpha_i-\beta_j)\\ &=n^n(-1)^{n(n-1)}\prod_{i,j}(\beta_j-\alpha_i)\\ &=n^n\prod_jf(\beta_j). \end{aligned} $$ We can calculate $$ f(\beta_j)=\beta_j(\beta_j^{n-1}-b)+c=c-(b-K^{n-1})\beta_j=c-\frac{b(n-1)K}n\zeta^j. $$ so $$ \operatorname{disc}(f)=n^n\prod_j(c-\frac{b(n-1)K}n\zeta^j). $$ We can use the factorization $$ u^{n-1}-v^{n-1}=\prod_j(u-\zeta^jv) $$ with $u=c$ and $v=b(n-1)K/n$ to evaluate this. The result is $$ \operatorname{disc}(f)=n^n\left(c^{n-1}-\left(\frac{b(n-1)K}n\right)^{n-1}\right) =n^n\left(c^{n-1}-\frac{(n-1)^{n-1}b^n}{n^n}\right), $$ which equals the claimed formula.