Apologies in advance if this turns out to be simple. So far I haven't found a proof or a reference.


Although I like $p$ to be a prime, I can ask the following for positive integers $n$ and $p$, using what should be clear notations for the $n$th cyclotomic polynomial and Euler's totient function: Given $p \gt 1$, is


$$ \mid \Phi\_n(p)/p^{\phi(n)} \mid \lt p/(p-1)$$
for every $n$?


Indeed, if $n$ is a power of a prime $q$, we have the left hand quantity bounded by $p^{n/q}/(p^{n/q} -1)$, and the lim sup over all primes $n$ achieves $p/(p-1)$. The case for composite $n$ is not clear to me, thus the question, but I would hope for a tighter bound (perhaps involving the smallest prime power factor of $n$) than $p/(p-1)$. 


An equivalent question asks to verify the bound
on $\Phi\_n(1/p)$. Of course the product of such quantities (to an appropriate power as $n$ runs over divisors of some $m$) will satisfy the bound, but this does not seem to help. If there is a reference offered that says (something like) the coefficients of cyclotomic polynomials grow slowly enough to exhibit the bound, I will read that. I am hoping for a simpler proof than that.


I am looking at (the moral equivalent of) prime factors of $\Phi\_n(p)$ and wanted to make sure these values aren't much bigger than I think they are.
I would be satisfied with a coarse bound (replace $p/(p-1)$ by $2$, say), but
I think much more can be said.


**UPDATE 2015.10.23** More now has been said, with my revised take on Jameson's presentation posted as a separate answer. For me, the key parts will be that $p \geq 2 $ prime and $p/(p-1)$ can be replaced by real $x \gt (2 - \epsilon)^{r/n}$ and $x^{n/r}/(x^{n/r} - 1)$, where $r=$rad$(n)$. Thanks again
all, and special thanks to Peter Mueller. **END UPDATE 2015.10.23**


**UPDATE 2015.10.21:** Thanks to Peter Mueller, I read from notes of G.J.O. Jameson at <http://www.maths.lancs.ac.uk/~jameson/cyp.pdf> on
cyclotomic polynomials of a sharper result, which indeed is simpler but also more challenging. I remove some of the challenge by interpreting some highlights here (hopefully without errors), but I recommend following the development of the notes as it proceeds in small but useful steps, with a certain degree of economy that takes ones breath away.


First, Jameson notes in 1.3 an inversion relation involving $\Phi\_n(1/x)$ 
and real,nonzero $x$ that appears below.
Jameson also prepares in 1.12 to work with squarefree indices through 
using $n\_0=$rad$(n)$ and the identity $\Phi\_n(x) = \Phi\_{n\_0}(x^{n/n\_0})$. I modify and sketch a strict inequality (Lemma 1.19) which is used:
For $0 \lt x \lt 1, m, a,\ldots,b$ positive integers (so also $0 \lt x^{powers} \leq x$),


\begin{eqnarray\*}
(1 -x^m)(1-x^{m+a})\ldots(1-x^{m+b}) & \geq &
(1 - x^m - x^{m+a} - \ldots - x^{m+b}) \\
& \gt & 1 - ( x^m + x^{m+1} + \ldots ) = 1 - x^m/(1-x) \\
\end{eqnarray\*}


Then Jameson has 1.20, which I rewrite and restrict to squarefree integers $n$, as one actually gets better bounds/ranges for when $n$ is not squarefree. 


1.20 (rewritten) Let $n>1$ be squarefree with $j=1$ if the number $k$ of distinct prime factors
of $n$ is an even number, and $j=-1$ if $k$ is odd. Let $0 \lt x \leq 1/2$. Then


$$1-x \lt \Phi\_n(x)^j \lt 1.$$


Note when $n$ is 1, one has $\Phi\_1(x)= x-1$ which is negative on the domain considered.


Using the inversion $\Phi\_n(x)/x^{\phi(n)} = \Phi\_n(1/x)$ when $n \gt 1$, this gives for $2 \leq x$
\begin{eqnarray\*}
(x-1)/x && \lt \Phi\_n(x)/x^{\phi(n)} \lt 1, && k=2m \\
1 && \lt \Phi\_n(x)/x^{\phi(n)} \lt x/(x-1), && k=2m+1 . \\
\end{eqnarray\*}


Using the relation for general non-squarefree indices, one can improve the $x$ in $1-x$ to $x$ to a fractional power, as well as extend the range a little. I am still working this part out. Even working out the statement using the inversion requires care. I think the results are both simple and challenging, and I am glad to share this on MathOverflow.


Jameson uses the tools carefully, working out the squarefree case in about half a page of elementary reasoning which I am still perusing. I am joyed. I'm also willing to buy Jameson two hot beverages. Peter Mueller can drop by 
and ask me for a toasted bagel.
**END UPDATE 2015.10.21**


Gerhard "Wants To Stop Spinning Head" Paseman, 2015.10.19



Cyclotomic polynomials: $\Phi_n(p)$ is like $p^{\phi(n)}$ for big enough $p$, right?
Using the Mobius inversion formula, one may write 


$$\Phi \_n(X)= \prod \_{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get 
$$\frac{\Phi \_n(p)}{p^{\phi (n)}}=\prod \_{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$ 


[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get 
$$\log \big (\frac{\Phi \_n(p)}{p^{\phi (n)}}\big )= \sum \_{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum \_{d\mid n} \mu (\frac{n}{d}) \sum \_{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate 
$$0\leq \mid \sum \_{d\mid n}\mu (\frac{n}{d})x^d \mid \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate  

$$\mid \log \big( \frac{\Phi \_n(p)}{p^{\phi (n)}}) \mid \leq \sum \_{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi \_n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies 
$$\frac{\Phi \_n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.


