Coefficient of $n$th cyclotomic polynomial equals $-\mu(n)$ For a polynomial $f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X]$ we define $\varphi(f):=a_1 \in \mathbb{Q}$. Now I want to show that for the $n$th cyclotomic polynomial $\Phi_n$ it holds that $$\varphi(\Phi_n)=-\mu(n)$$
where $\mu(n)$ is the Möbius function. What I know is that $\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}$.
 A: Note that the Möbius function $\mu(n)$ can be defined as the unique
arithmetic function $f$ that fulfills
$$ \forall n\in\mathbb{N}: \sum_{d\mid n} f(d) =
\begin{cases}1 & \text{for $n=1$}\\ 0 &\text{for $n>1$}\end{cases}\tag{1}$$
Now define
$$g(d) = \left[\Phi_d(X)\right]_{-1}$$
where $[\cdot]_{-1}$ is the next-to-leading coefficient.
It follows that
$$\sum_{d\mid n} g(d) = \sum_{d\mid n}\left[\Phi_d(X)\right]_{-1}
= \left[\prod_{d\mid n} \Phi_d(X)\right]_{-1}
= \left[X^n-1\right]_{-1}
= \begin{cases}
-1 & \text{for $n=1$}\\ 0 &\text{for $n>1$}
\end{cases}$$
And by the above definition of $\mu(n)$ you can conclude $g(n)=-\mu(n)$,
which implies $\left[\Phi_n(X)\right]_{-1}=-\mu(n)$.
Update: For monic univariate polynomials $f_1,f_2$ of degree at least $1$
we have
$$\left[f_1(X)\,f_2(X)\right]_{-1} = \left[f_1(X)\right]_{-1}+\left[f_2(X)\right]_{-1}$$
because the coefficient in question is the negative sum of the roots
of $f_1$ and $f_2$ (with multiplicity).
Even without considering roots,
this follows from looking at how the product expands.
Update: Proof
that $\mu$ is the unique arithmetic function with property $(1)$:
For $n=1$ we obtain $f(1)=1=\mu(1)$.
For $n>1$, let $n=p_1^{e_1}\cdots p_r^{e_r}$ where $p_1,\ldots,p_r$
are pairwise distinct primes and $e_1,\ldots,e_r$ are positive integers.
Then
$$\begin{align}
\sum_{d\mid n} \mu(d) &= \sum_{j_1=0}^{e_1}\cdots\sum_{j_r=0}^{e_r}
\mu(p_1^{j_1}\cdots p_r^{j_r}) = \sum_{j_1=0}^{e_1}\cdots\sum_{j_r=0}^{e_r}
\begin{cases} 0 & \text{if any $j_i>1$}\\
(-1)^{j_1+\cdots+j_r} & \text{otherwise}\end{cases}\\
&= \left(\sum_{j_1=0}^{1}(-1)^{j_1}\right)\cdots
\left(\sum_{j_r=0}^{1}(-1)^{j_r}\right) = 0
\end{align}$$
Thus $\mu$ has property $(1)$.
Now for uniqueness:
Let $\mu_1,\mu_2$ be arithmetic functions with property $(1)$.
Then necessarily $\sum_{d\mid n}\mu_1(d)=\sum_{d\mid n}\mu_2(d)$
for all positive integers $n$.
Suppose $\mu_1\neq \mu_2$, then there exists a minimal positive integer $m$
such that $\mu_1(m)\neq \mu_2(m)$. But then
$\sum_{d\mid m}\mu_1(d)\neq\sum_{d\mid m}\mu_2(d)$ which contradicts the
hypothesis. (All $d<m$ do not make any difference since $m$ is minimal
in that respect, but $d=m$ does make a difference.)
The only remaining possibility is $\mu_1=\mu_2$.
A: Working out the following definition of the Cyclotomic Polynomial
$$    {\displaystyle \Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(x-e^{2i\pi {\frac {k}{n}}}\right),}
$$
you'll get
$$
{\displaystyle x^{\varphi(n)}+x^{\varphi(n)-1} \left({ -\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} \right)} + \dots + 1,
$$
because every coefficient of the expanded polynomial may be represented as an elementary symmetric polynomial, which is $e_1(\cdot)$ in your case.
Using the defintion here, you'll spot right away that this gives the Möbius function:
$$    {\displaystyle -\mu (n)=-\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}}
$$
A: Hopefully more understandable for beginners:
The Möbius function $\mu$ is characterized by:
$$\forall n\in\Bbb N\quad\sum_{d\mid n}\mu(d)=\begin{cases}1&\mbox{ if } n=1\\ 0&\mbox{ if } n>1.\end{cases}$$
Let $S(n)$ denote the sum of complex primitive $n$-th roots of $1.$ In order to prove that $S=\mu,$ all we have to do is to prove that $S$ satisfies the same equation.
Now, $\sum_{d\mid n}S(d)$ is the sum of all $n$-th roots of $1:$
$$\sum_{d\mid n}S(d)=\sum_{k=0}^{n-1}(e^{2\pi i/n})^k=\begin{cases}1&\mbox{ if } n=1\\ \frac{1-(e^{2\pi i/n})^n}{1-e^{2\pi i/n}}=0&\mbox{ if } n>1,\end{cases}$$
and we are done.
