Combinatorial interpretation of this identity of Gauss? Gauss came up with some bizarre identities, namely
$$
\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}.
$$
How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.

The typical analytic proof is not difficult and is an easy consequence of Jacobi's triple product $$\sum_{n=-\infty} ^{\infty} z^{n} q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$ for all $z, q$ with $z\neq 0,|q|<1$. Let's put $z=-1$ to get the sum in question. The corresponding product is equal to $$\prod(1-q^{2n})(1-q^{2n-1})^{2}=\prod(1-q^{n})(1-q^{2n-1})=\prod \frac{(1-q^{2n})(1-q^{2n-1})} {1+q^{n}}=\prod\frac{1-q^{n}} {1+q^{n}}$$ which completes the proof. 
The proof for Jacobi's triple product is non-trivial / non-obvious and you may have a look at the original proof by Jacobi in this blog post.

On the other hand Franklin obtained a nice and easy combinatorial proof of the Euler's Pentagonal theorem which is equivalent to Jacobi Triple Product. 
