A strange sum over bipartite graphs While mucking around with some generating functions related to enumeration of regular bipartite graphs, I stumbled across the following cutie.  I wonder if anyone has seen it before, and/or if anyone sees a nice interpretation.  The sum is over all simple bipartite graphs $G$ with $n$ vertices on each side, the product is over all $2n$ vertices $v$ of $G$ and $d_G(v)$ is the degree of vertex $v$ in graph $G$.
$$\sum_{G\subseteq K_{n,n}} ~ \prod_{v\in V(G)} (n-2d_G(v)) = 2^{n^2}n!~.\kern 3cm (1)$$
Addition 1, proof.  Consider $n^2$ commuting indeterminates $\{x_{ij}\}_{i,j=1\ldots n}$. The polynomial
$$P(\boldsymbol{x}) = \prod_{i=1}^n \sum_{j=1}^n x_{i,j} \times \prod_{j=1}^n \sum_{i=1}^n x_{i,j}.$$
has terms with each variable having power 0, 1 or 2. Consider the total coefficient $C~$ of the terms with only even powers.  One way is to sum each variable over $\pm 1$, which makes the terms we don't want cancel out and the others get multiplied by $2^{n^2}$. This gives (1), interpreting $G$ as the bipartite graph whose edges are the variables with value $-1$.  Alternatively, note that each term has total degree $2n$ and the only possible such term with even degrees is $\prod_{i=1}^n x_{i,\sigma(i)}^2$ for some permutation $\sigma$.  This shows $C=n!~$.
Addition 2, generalisation. Define the numbers
$$\rho(n,k,d) = \sum_{j\ge 0} (-1)^j \binom{d}{j} \binom{n-d}{k-j}.$$
Let $k_1,\ldots,k_{2n}$ be a sequence of nonnegative integers.  Then
$$\sum_{G\subseteq K_{n,n}} ~ \prod_{v\in V(G)} \rho(n,k_v,d_G(v)) 
  = 2^{n^2} B(\boldsymbol{k}), $$
where $B(\boldsymbol{k})$ is the number of simple bipartite graphs with vertex $v$ having degree $k_v$ for all $v$.  The case (1) follows from $\rho(n,1,d)=n-2d$. The proof of the general case is similar.
 A: I'm not sure if this is the right interpretation or not...it may really just be another way of encoding the generating function argument.  Let $H$ be a random bipartite graph where every edge appears independently with probability $1/2$.  Then the left hand side is 
$$2^{n^2} E \left(\prod_v f(v) \right),$$
where $f(v)$ is equal to $\sum_u x(u,v)$ and $x(u,v)$ is $1$ if an edge is not present, $-1$ if an edge is present.  Expanding out the product and using linearity of expectation, we can write this as 
$$2^{n^2} \sum_{\sigma} E \left(\prod_{v} x(v,\sigma(v))\right)$$
Where $\sigma$ consists of all mappings taking each vertex to a vertex on the opposite side.  
Any $\sigma$ for which some edge $(v,\sigma(v))$ appears only once has $0$ expectation due to independence.  The $\sigma$ for which every edge appears for both of its endpoints correspond to matchings between the left and right side, of which there are $n!$.  (This last observation corresponds to the fact that the expected square of the permanent of an $n \times n$ random Bernoulli matrix is $n!$...I think it goes at least back to Turan).  
