number theory - Help finishing proof with polynomial discriminant?

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