Prove that the discriminant of $$f(x) = x^n + nx^{n-1} + n(n-1)x^{n-2} + \cdots + n(n-1)\ldots (3)(2)x + n!$$ is $(-1)^{n(n-1)/2}(n!)^n$.

So far, I let $\alpha_1,\ldots, \alpha_n$ be the roots of $f(x)$. Taking the derivative of $\log f(x) = \sum_{i=1}^n \log(x - \alpha_i)$, we have that $$\frac{f'(x)}{f(x)} = \sum_{i=1} \frac{1}{x-\alpha_i}.$$ Thus, $$f'(\alpha_j) = \prod_{i=1, i\neq j}^n (\alpha_j - \alpha_i)$$ Then the discriminant is $$D = \prod_{i<j} (\alpha_i - \alpha_j)^2 = (-1)^{n(n-1)/2}\prod_{i\neq j} (\alpha_i - \alpha_j) = (-1)^{n(n-1)/2}\prod_{k=1}^n f'(\alpha_k).$$


Note that $f'(x) = f(x) - x^n$. This implies that $f'(\alpha_k) = -\alpha_k^n$.

Your product then simplifies via $$ \prod_{k=1}^n f'(\alpha_k) = \prod_{k=1}^n -\alpha_k^n = (-1)^n \left(\prod_{k=1}^n \alpha_k\right)^n = (-1)^n \left((-1)^nn!\right)^n = (n!)^n $$

Here we use the fact that the product of the roots of $f(x)$ is equal to $(-1)^n$ times the constant term $n!$.


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