Transforming numbers of irreducible polynomials Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number,
stand for the number of irreducible monic polynomials  of degree $n$ in the
polynomial ring $\mathbf{F}_{q}[X]$ over the finite field
$\mathbf{F}_{q}$. Since
\begin{equation*}
  M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d}
\end{equation*}
we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$
be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of
$n = n_1e_1 + \cdots + n_se_s$.  Consider the sum of polynomials in
$\mathbf{Q}[q]$
\begin{equation*}
  \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} 
  \prod_{i=1}^s \binom{M(n_i)}{e_i} 
\end{equation*}
ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and
it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all
$n>1$. Is this true?
 A: Yes, it is true. Your expression is the coefficient of $x^n$ in the following product:
$$\prod_{P\text{ monic irreducible}} (1-x^{\deg P}) = \prod_{n} (1-x^n)^{M(n)}.$$
The Zeta function of $\mathbb{F}_q[X]$ is
$$\sum_{f \text{ monic}} x^{\deg f} = \sum_{n \ge 0} q^n x^n = \frac{1}{1-qx}.$$
The Euler product identity tells us that
$$\frac{1}{1-qx} = \prod_{P\text{ monic irreducible}} (1-x^{\deg P})^{-1} =\prod_{n} (1-x^n)^{-M(n)}.$$
Taking its reciprocal, we find that
$$1-qx = \prod_{P\text{ monic irreducible}} (1-x^{\deg P}).$$
Now it is just a matter of comparing coefficients on both sides.

Interpretation: Let $\mu: \mathbb{F}_q[X] \to \mathbb{C}$ be the polynomial Möbius function, defined by
$$\mu(f) = \begin{cases} 0 & f \text{ not squarefree} \\ (-1)^k & f=p_1 \cdots p_k (p_i \text{ distinct irreducibles}) \end{cases}.$$ 
The term $\prod_{i=1}^{s} \binom{M(n_i)}{e_i}$ counts the number of monic, squarefree polynomials of degree $n$ whose factorization consists of $e_i$ irreducibles of degree $n_i$. The polynomial Möbius function $\mu(\bullet)$ assumes the value $(-1)^{\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as
$$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$
Since $\mu(f)=0$ for $f$ which is not squarefree, your claim is the same as
$$n>1 \implies \sum_{f \text{ monic of degree }n} \mu(f) = 0.$$
This is classical, and due to L. Carlitz, whose proof was exactly as above. It should be compared with the following equivalent formulation of the Riemann Hypothesis:
$$\sum_{n \le x } \mu(n) = O(x^{\frac{1}{2}+\varepsilon}),$$
where this time $\mu$ is the integer Möbius function.

It is interesting to ask whether $\sum_{f\text { monic of degree }n} \mu(f)$ may be evaluated without the use of formal power series and the Euler product identity. I will present such a way. Let $M_q$ denote the set of monics in $\mathbb{F}_q[X]$. Given two functions $\alpha, \beta : M_q \to \mathbb{C}$, one defines their "Dirichlet convolution" as the following function:
$$(\alpha*\beta) (f) = \sum_{d_1 \cdot d_2 = f} \alpha(d_1) \beta(d_2).$$
Let $\zeta$ denote the constant function $1$ on $M_q$. Let $\delta_{1}$ denote the indicator function of the constant polynomial $1$. The "defining" property of $\mu$ is
$$\mu * \zeta = \delta_1.$$
By summing the above over all monic polynomials of degree $n$ (>0) and changing the order of summation, we get that
$$\sum_{\deg d \le n} \mu(d) q^{n-\deg d} = 0,$$
or equivalently
$$\sum_{\deg d \le n} \frac{\mu(d)}{q^{\deg d}} =0.$$
Plugging $n=m,m+1$ in the above and subtracting, we get that
$$m>0 \implies \sum_{\deg d = m+1} \mu(d) = 0.$$
A: Here is a reformulation of Ofir Gorodetsky's excellent answer to my question:
Let $(a(n))_{n \geq 1}$ be a sequence of rational numbers. Define
the transformed sequence $T(a)$ of $a$ to have $n$th element
\begin{equation*}
  T(a)(n) = \sum_{n_1^{e_1} \cdots n_s^{e_s} \in P(n)} (-1)^{\sum e_i}
    \prod_{i=1}^s \binom{A(n_i)}{e_i}, \qquad n \geq 1
\end{equation*}
Then
\begin{equation*}
  1+ \sum_{n \geq 1} T(a)(n)x^n = \prod_{n \geq 1} (1-x^n)^{a(n)}
\end{equation*}
We now apply this to the sequence $M(n)(q)$. From the classical formula
\begin{equation*}
  \frac{1}{1-qx} = \prod_{n \geq 1} (1-x^n)^{-M(n)(q)}
\end{equation*}
we get that
\begin{equation*}
   1+ \sum_{n \geq 1} T(M(n)(q))x^n   =  \prod_{n \geq 1}
  (1-x^n)^{M(n)(q)} = 1-qx
\end{equation*}
This shows that $T(M)(1)=-q$ and $T(M)(n)=0$ for all $n>1$.
