Let $n \geq 2$, and $\alpha$ be an algebraic integer with minimal polynomial $$f(x) = x^n + ax + b, \, \, \, a, b \in \mathbb{Z}.$$ Show that $$\Delta(\alpha) = (-1)^{\frac{n(n-1)}{2}}((-1)^{n-1}(n - 1)^{n -1}a^n + n^n b^{n-1} ).$$
** So since I know that $\Delta(\alpha) = (-1)^{\frac{n(n-1)}{2}} \prod_{j = 1}^n f'(\alpha_j )$, I just need to show $\prod_{j = 1}^n f'(\alpha_j ) = (-1)^{n-1} (n - 1)^{n -1} a^n + n^n b^{n-1}$.
So for practice I first tried for $n =2$ and got $\alpha$ has minimal polynomial, $$x^2 + ax + b = (x - \alpha_1)(x - \alpha_2) = x^2 -(\alpha_1 + \alpha_2)x + \alpha_1\alpha_2 \, \, \, \, \, \, \, \, \, \, (1).$$ So $f'(\alpha_1)f'(\alpha_2) = (2\alpha_1^{2 - 1} + a)(2\alpha_2^{2 - 1} + a) = (2\alpha_1 + a)(2\alpha_2 + a)$ $= 4\alpha_1\alpha_2 + 2a(\alpha_1 + \alpha_2) + a^2 = 4b + 2a(\alpha_1 + \alpha_2) + a^2 = 4b + 2a(-a) + a^2$ $= 4b -2a^2 + a^2 = -a^2 + 4b = (-1)^{2-1} (2 - 1)^{2 -1} a^2 + 2^2 b^{2-1}$, where equalities $4$ and $5$ follow from $(1)$.
Then I tried for $n = 3$ and got $\alpha$ has minimal polynomial, $$x^3 + ax + b = (x - \alpha_1)(x - \alpha_2)(x - \alpha_3)$$ $$= x^3 -(\alpha_1 + \alpha_2 + \alpha_3)x^2 + (\alpha_1\alpha_2+ \alpha_1\alpha_3 + \alpha_2 \alpha_3)x - \alpha_1\alpha_2\alpha_3 \, \, \, \, \, \, \, \, \, \, (2).$$ So $f'(\alpha_1)f'(\alpha_2)f'(\alpha_3) = (3\alpha_1^{3 - 1} + a)(3\alpha_2^{3 - 1} + a)(3\alpha_3^{3 - 1} + a)$ $= (3\alpha_1^2 + a)(3\alpha_2^2 + a)(3\alpha_3^2 + a)$ $= 27\alpha_1^2\alpha_2^2\alpha_3^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2) + a^3$ $= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2) + a^3$ $= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 0) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$ $= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 2(0)\alpha_1\alpha_2\alpha_3) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$ $= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 2(\alpha_1 + \alpha_2 + \alpha_3)\alpha_1\alpha_2\alpha_3) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$ $= 27b^2 + 9a(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 )^2 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$ $= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$ $= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - 0 ) + a^3$ $= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - 0^2 ) + a^3$ $= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - (\alpha_1 + \alpha_2 + \alpha_3)^2) + a^3$ $= 27b^2 + 9a^3 + 3a^2(- 2(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3) ) + a^3$ $= 27b^2 + 9a^3 + 3a^2(- 2(a) ) + a^3$ $= 27b^2 + 9a^3 - 6a^3 + a^3 = 27b^2 + 4a^3$ $= (-1)^{3-1} (3 - 1)^{3 -1} a^3 + 3^3 b^{3-1}$, where equalities $4$, $7$, $9$, $12$, and $14$ follow from $(2)$.
**At this point I was hoping I was ready for the general case so I started off with Now suppose for some $n \geq 2$, and $\alpha$ has minimal polynomial, $$x^n + ax + b = (x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_n)$$ $$= x^n - \sum_{i = 1}^n \alpha_ix^{n -1} + \sum_{i = 1}^{n - 1} \sum_{j > i}^n \alpha_i \alpha_jx^{n -2} - \sum_{i = 1}^{n - 2}\sum_{j > i}^{n - 1}\sum_{k > j}^n \alpha_i\alpha_j \alpha_k x^{n - 3} +$$ $$\cdots + (-1)^{n - 1}\sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n\alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_{n-1}}x + (-1)^n\alpha_1\alpha_2\cdots\alpha_n \, \, \, \, \, (3).$$
So $\prod_{j = 1}^n f'(\alpha_j ) = (n\alpha_1^{n-1} + a)(n\alpha_2^{n - 1} + a)\cdots (n\alpha_n^{n - 1} + a)$
$= n\alpha_1^{n - 1}n\alpha_2^{n - 1}\cdots n\alpha_n^{n-1} + \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n - 1} \cdots n\alpha_{i_{n - 1}}^{n - 1}a$ $+ \sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n - 1} \cdots n\alpha_{i_{n - 2}}^{n - 1}a^2 +$ $ \cdots + \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n-1} \cdots \alpha_{i_{n + 1 - j}}^{n - 1}a^{j - 1} +$ $\cdots + \sum_{i = 1}^n n\alpha_i^{n - 1}a^{n - 1} + a^n$
$= n^n(\alpha_1\alpha_2\cdots\alpha_n)^{n - 1} + n^{n - 1}a \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 1}})^{n - 1}$
$+ n^{n - 2}a^2\sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 2}})^{n - 1} +$ $\cdots + n^{n + 1 - j} a^{j - 1}\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} +$ $\cdots + na^{n - 1}\sum_{i = 1}^n \alpha_i^{n - 1} + a^n$
$ = n^nb^{n - 1} + n^{n - 1}a \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 1}})^{n - 1}$ $+ n^{n - 2}a^2\sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 2}})^{n - 1} +$ $\cdots + n^{n + 1 - j} a^{j - 1}\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} +$ $\cdots + na^{n - 1}\sum_{i = 1}^n \alpha_i^{n - 1} + a^n$
Now I am pretty sure that the idea is to use the fact that from $(3)$ we have $(-1)^{n - 1}\sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n\alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_{n-1}} = a$ and the coefficients in front of $x^{n-1}, x^{n - 2}, \cdots, x^2$ are zero, to some how rewrite $ \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1}$ as a bunch of products and sums of the coefficients on the right hand side of $(3)$ to get in such a way to get $ \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} = (-1)^{n -1}(_{n - 1}$C$_{j - 2})(-1)^{j - 2)a^{n + 1 - j}$, and then use the binomial theorem to finish.
The problem is every time I try to play with reweighting $\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1}$ I get an algebraic nightmare
Let us start from the formula which you use, $\Delta(\alpha)=(-1)^{n(n-1)/2}N(f'(\alpha))$, where $N$ is the norm map of $\mathbf Q(\alpha)/\mathbf Q$ (note that you don't even need $\alpha$ to be an algebraic integer); this comes simply from the calculation of a Vandermonde determinant, see e.g. D. Marcus' "Number Fields", chap. 2, thm. 8. In the particular case of a trinomial $f(X)=X^n +aX+b$, put $\beta=f'(\alpha)=n\alpha^{n-1}+a=-(n-1)a-nb\alpha^{-1}$ because $f(\alpha)=0$, so that $\alpha= -nb(\beta+(n-1)a)^{-1}$. The minimal polynomial of $\beta$ is the numerator of $f(-nb(Y+(n-1)a)^{-1}))$, which is (all calculations done), $(Y+(n-1)a)^n-na(Y+(n-1)a)^{n-1}+(-1)^nb^{n-1}$. The norm of $\beta$ is $(-1)^n$ times the constant term of this polynomial, i.e. $n^n b^{n-1}+(-1)^{n-1}(n-1)^{n-1} a^n$, and this gives the desired formula for $\Delta(\alpha)$. You can check that for $n=2$ (resp. $3$), you recover the usual expressions $a^2-4b$ (resp. $-(4a^3+27b^2)$) .