Peter wrote: If $p$ is a prime and $n$ an integer such that $1<n \le p$, then \[\phi \left( \sum_{k=0}^{p-1}n^{k}\right) \equiv 0 \; \pmod{p}.\] At first let's prove that $gcd(\sum_{k=0}^{p-1}n^{k};n-1)=1$. \[\sum_{k=0}^{p-1}n^{k}=\sum_{k=0}^{p-1}(n^{k}-1)+p\] then \[gcd(\sum_{k=0}^{p-1}n^{k};n-1)=gcd(\sum_{k=0}^{p-1}(n^{k}-1)+p;n-1)=gcd(p;n-1)=1\] because $n-1<p$. Now we have that $\phi(n^{p}-1)=\phi(\sum_{k=0}^{p-1}n^{k})\phi(n-1)$. From Euler's theorem we have that $(n^{p}-1)|n^{\phi(n^{p}-1)}-1$ then $p|\phi(n^{p}-1)$. Therefore, to finish the proof we need to show that $(p;\phi(n-1))=1$. But latter is obvious from the condition $n-1<p$ and the formula \[\phi(l)=p_{1}^{\alpha_{1}-1}...p_{r}^{\alpha_{r}-1}({p_{1}}-1)...(p_{r}-1), \text{for}l=p_{1}^{\alpha_{1}}...p_{r}^{\alpha_{r}}\]

Peter wrote: If $ p$ is a prime and $ n$ an integer such that $ 1 < n\le p$, then \[ \phi\left(\sum_{k=0}^{p-1}n^{k}\right)\equiv 0\;\pmod{p}.\] Note that $ 1+n+\cdots+n^{p-1}=\frac{n^{p}-1}{n-1}$. Here are the steps. Step 1. $ p\;\vert\;\phi( n^{p}-1 )$. Step 2. $ \gcd\left(\frac{n^{p}-1}{n-1}, n-1\right)=1$. Step 3. $ p\not\vert\;\phi(n-1)$. From Step 1 and Step 2, we obtain \[ p\;\vert\;\phi\left( n^{p}-1\right) =\phi\left(\frac{n^{p}-1}{n-1}\right)\phi(n-1).\] Since $ p$ is prime, it follows from Step 3 that $ p\;\vert\;\phi\left(\frac{n^{p}-1}{n-1}\right)$. Proof of Step 1. It's an easy job to prove that $ n$ has order $ p$ modulo $ n^{p}-1$. Since $ n^{\phi( n^{p}-1)}\equiv 1 (mod\; n^{p}-1)$, we obtain $ p\;\vert\;\phi( n^{p}-1 )$. Proof of Step 2. We offer two ways. First Way. Observe that $ n^{k}\equiv 1\; (mod\; n-1)$ for all $ k$. It follows that \[ \frac{n^{p}-1}{n-1}\equiv\sum_{k = 0}^{p-1}n^{k}\equiv\sum_{k = 0}^{p-1}1\equiv p\; (mod\; n-1)\] so that \[ \gcd\left(\frac{n^{p}-1}{n-1}, n-1\right) =\gcd\left( p, n-1\right).\] Since $ p$ is prime and since $ n-1 < p$, we have $ \gcd\left( p, n-1\right)=1$. Second Way. Assume that $ \gcd\left(\frac{n^{p}-1}{n-1}, n-1\right)>1$. Let $ q$ be a common prime divisor of $ \frac{n^{p}-1}{n-1}$ and $ n-1$. Since $ n\equiv 1\; (mod\; q)$, we have \[ \frac{n^{p}-1}{n-1}\equiv\sum_{k=0}^{p-1}n^{k}\equiv\sum_{k=0}^{p-1}1^{k}\equiv p\; (mod\; q).\] Since $ \frac{n^{p}-1}{n-1}$ is divisible by $ q$ and since both $ p$ and $ q$ are primes, this implies that $ q=p$. Since $ q$ divides $ n-1$, we obtain $ n-1\geq q$. However, we have $ p\geq n > n-1\geq q=p$, which is a contradiction. Proof of Step 3. Suppose that $ p\;\vert\;\phi(n-1)$. Clearly, $ n-1>1$. Now, we write \[ n-1 =\prod_{j=1}^{k}{q_{j}}^{e_{j}},\] where $ q_{1},\cdots, q_{j}$ are distinct primes and $ e_{1},\cdots, e_{j}$ are positive integers. It follows that \[ \phi(n-1)=\prod_{j=1}^{k}{q_{j}}^{e_{j}-1}\left( q_{j}-1\right).\] Since $ p\;\vert\;\phi(n-1)$ and since $ p$ is prime, this means that $ p\;\vert\; q_{j}$ or $ p\;\vert\; q_{j}-1$ for some $ j\in\{1,\cdots, k\}$. In any case, we get $ p\leq q_{j}$. It follows that $ p\leq q_{j}\leq n-1$. This is a contradiction for $ n\leq p$.

Actually, the condition $ n\leq p$ is not neccessary. Lemma (well known): Every prime divisor of $ \frac{n^p-1}{n-1}$ is either equal to $ p$ or congruent to $ 1$ modulo $ p$. Suppose now that $ \frac{n^p-1}{n-1}$ has a prime divisor $ q\equiv 1\pmod p$. Then it follows from $ q-1|\phi(\frac{n^p-1}{n-1})$ that $ p|\phi(\frac{n^p-1}{n-1})$. If $ \frac{n^p-1}{n-1}=p^k$ for some positive integer $ k$ then we only need to prove that $ k>1$. But this immediately follows from Bernoullis inequality.