Combinatorial sum in a problem with a Fermi gas I'm solving a  problem involving a Fermi gas. There is a specific sum I cannot figure my way around. 
A set of equidistant levels, indexed by $m=0,1,2 \ldots$,  is populated by spinless fermions with population numbers $\nu_m =0 $ or $1$. I need to compute the following sum over the set of all possible configurations $\{  \nu_l \}$:

$Q(\beta,\beta_c) = \sum_{\{  \nu_l \}} \sum_{l} \prod_m \exp({\beta_c \, l \, \nu_l}-{ [ \beta \, m + i \phi] \, \nu_m} )$.

Any hints on how to deal with this are appreciated. This is not homework, it is a research problem.
It is known that $\beta >0$,  $\beta_c>0$, and $\phi \in [0; 2 \pi ]$.
EDIT: corrected with the complex phase (the sum is coming from a generating function)
 A: This is not an answer just some thoughts from playing with the expression. I've read the question before you included the phase, so for now let $\phi = 0$ (sorry it this makes my response useless for you).
I'll simply write $Z$ instead of your $Q(\beta, \beta_c)$ and also drop the arguments where obvious. Denote by $Z_{abc\dots}$ the partition function where we do not include the sites at $a, b, c, \dots$ in the problem. Also denote $f_k = 1 + \exp(-\beta k)$ and $g_k = 1 + \exp((\beta_c - \beta) k)$.
Now (unless I screwed up), by summing over the site at $k$ we can get the relation
$$Z = f_kZ_k + g_k \sum_{\nu \setminus k} \prod_{m \neq k} \exp(-\beta m \nu_m) $$
and iterating it
$$Z = \left( \prod_{m \in abc\dots z} f_m \right) Z_{abc\dots z} + $$
$$ \left(g_a f_b \dots f_z + f_a g_b \dots f_z + \cdots + f_a f_b \dots g_z)  \right) \sum_{\nu \setminus abc\dots z} \prod_{m \neq abc \dots z} \exp(-\beta m \nu_m).$$
It is a simple observation that for the reduces system consisting of a single level $a$ we get $Z_{bc \ldots z} = g_a$ so the first term above gives a similar contribution like the other terms (all but one factors are $f$ and one of them is $g$). Therefore, we can write
$$Z = \left( \prod_{m} f_m \right) \left ( \sum_k \frac{g_k} { f_k} \right).$$
These expressions are exact in case we have finite number of states. Otherwise they are just formal and are to be understood as limits only if everything converges.
A: The answer by Marek was so useful for me, that I wish to share the full problem and the resulting answer (to improve the lasting value of this Q-A):
Problem: In an equilibrium Fermi gas at inverse thermodynamic temperature 
$\beta$ defined by a set of single particle levels $\epsilon_m$ ($m=0,1,\ldots$)
which are populated by $n$ fermions, 
the canonical averages of an arbitrary single-particle field $\langle h_m(\nu_m) \rangle_n$
(where $\nu_m =0, 1$ is the occupation number) can be computed via the generating 
function
$ Z[h_m; z] \equiv \sum_{n=0}^{\infty} z^{n} \langle h_m(\nu_m) \rangle_n Z_n 
=\sum_{ \{ \nu_k\} } \sum_l h_l(\nu_l) \prod_m e^{-\beta \epsilon_m \nu_m} z^{\nu_m} $.
Solution:As was shown by Marek,
$Z[h_m; z]= Z(z) \sum_k \frac{g_k}{f_k}$
where 
$f_k  = 1 + e^{-\beta \epsilon_k} z$,  
$g_k  = h_k(0) + h_k(1) e^{-\beta \epsilon_k} z$ 
and
$ Z(z) \equiv \prod_m f_m$. 
The canonical partition functions $Z_n$
are generated by $Z(z)= \sum_{n=0}^{\infty} Z_n z^n$. 
