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It is well known that in the cyclotomic ring $\Bbb{Z}[\zeta_p]$ one has the equality of ideals $$(1-\zeta_p)^{p-1}=(p).$$ I'm struggling to find 'manageable' expressions for the associated units $$\frac{(1-\zeta_p)^{p-1}}{p}\qquad\text{ and }\qquad\frac{p}{(1-\zeta_p)^{p-1}},$$ on the power basis $\zeta_p^0,\ldots,\zeta_p^{p-2}$ of $\Bbb{Z}[\zeta_p]$. One approach I tried uses that $$(1-\zeta_p)^{p-1}=\prod_{n=1}^{p-1}(1-\zeta_p^n)\frac{1-\zeta_p^n}{1-\zeta_p}=\Phi_p(1)\prod_{n=1}^{p-1}\frac{1-\zeta_p^n}{1-\zeta_p}=p\cdot\prod_{n=1}^{p-1}\sum_{m=0}^{n-1}\zeta_p^m,$$ which gives the (to me) unmanageable expression $$\frac{(1-\zeta_p)^{p-1}}{p}=\prod_{n=1}^{p-1}\sum_{m=0}^{n-1}\zeta_p^m.$$ I've written this out explicitly for $p\leq7$, but I don't see a pattern yet. Are there nice expressions for the associated units above?

EDIT: I like to clarify that I'm looking for explicit expressions for the coefficients of these units on the standard power basis for $\Bbb{Z}[\zeta_p]$. And although the answer by sharding4 below does exactly that, I was hoping for expressions without denominators. That is, expressions from which it is immediately clear that this is an element of $\Bbb{Z}[\zeta_p]$.

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A fairly "manageable" expression in terms of the power basis can be obtained simply by applying the binomial theorem to $(1-\zeta_p)^{p-1}$ and using the fact that $\zeta_p^{p-1}=-\zeta_p^{p-2}-\zeta_p^{p-3}-\cdots - \zeta^2-\zeta - 1$. Then $$ \frac{(1-\zeta_p)^{p-1}}{p}=\sum_{k=1}^{p-2}\frac{(-1)^k\binom{p-1}{k}-1}{p}\zeta_p^{k} $$

sharding4
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  • This seems very nice, but how are the coefficients $\frac{(-1)^k\binom{p-1}{k}-1}{p}$ integers? I was hoping for non-fractional (polynomial?) expressions for the coefficients, which is why I took such a detour in my question. Perhaps I should clarify my question, or make it more specific. –  Jul 16 '17 at 17:51
  • @user38292 It isn't immediately obvious but $\frac{(-1)^k\binom{p-1}{k}-1}{p}$ is in fact an integer for $1\leq k\leq p-2$ – sharding4 Jul 16 '17 at 17:59
  • I know it is, for example by Pascal's rule for binomial coefficients, but I was hoping for expressions from which this is immediately obvious. I will try to work from your answer myself as well, thanks for the pointer. –  Jul 16 '17 at 18:01
  • Also, is there a similarly nice expression for the inverse? –  Jul 17 '17 at 03:37