I have the following polynomial: $$f(x)=\prod_{i\in(\mathbb{Z}/n\mathbb{Z})^\times} \left((x+\zeta_n^i)^n-(n+1)\right)$$ Here, $\zeta_n = e^{2\pi i/n}$ is a primitive $n$th root of unity.
I want a "closed form" for it, meaning: $$f(x) = \sum_{i =0}^{\varphi(n)n} a_i x^i$$ Where $\varphi(n)n$ is the degree of $f$. I've tried what seems "standard", in applying binomial theorem:
$$f(x) = \prod_{i\in(\mathbb{Z}/n\mathbb{Z})^\times}\left(\sum_{j = 0}^n\binom{n}{j}x^j(\zeta_n^i)^{n-k}-(n+1)\right)$$ It seems like the next step is probably to apply: $$\prod_{i = 0}^n\sum_{j = 0}^m a_{i,j} = \sum_{e\in\{1,\dots,m\}^n}\prod_{i = 0}^n a_{i,e_i}$$ This seems quite difficult though. Can someone suggest how to move forward with this?
For those interested, $f$ is the defining polynomial of $\mathbb{Q}(\zeta_n,(n+1)^{1/n})$ as a number field, which I'm interested in doing Sage computations in quickly. Hopefully, an explicit form of $f$ will help this.
Since if you can find $f^{(k)}(0)$ for all $k$, you can get the polynomial, a possibility would be to get these derivatives.
One way might be to apply the product rule for differentiating.
If $g(x) =\prod_{k=1}^m f_m(x) $ then $\ln g(x) =\sum_{k=1}^m \ln f_m(x) $ so $(\ln g(x))' =\dfrac{g'(x)}{g(x)} =\sum_{k=1}^m \dfrac{ f_m'(x)}{f_m(x)} $ so $g'(x) =g(x)\sum_{k=1}^m \dfrac{ f_m'(x)}{f_m(x)} $.
Another possibility is to look up "nth derivative of composite function" or "faa di bruno's rule" and see if these can help.