Gaussian for Grassmann variables Let $(\theta,A\theta)=\theta_i A_{ij}\theta_j$ where $A$ is some $(2\times2)$ antisymmetric matrix. 
I want to generalize the following 
$$I(A) =\int d\theta_1d\theta_2~ \exp\Bigg[\frac{1}{2}(\theta,A\theta)\Bigg]=\int d\theta_1d\theta_2~ (1+\theta_1\theta_2A_{12}) = A_{12}=\sqrt{\det A}$$
to the $n$-tuple case. 
Let now $$A:=\begin{bmatrix}
0    & 1       & \;     & \;    \\
\;-1    & 0     &  & \;    \\     
\;     & \;      & 0 & 1     \\
\;     & \;      & -1 & 0     \\
\;     & \;      & \;     & \, &\ddots   \\
\end{bmatrix}.$$
I evaluate, I get the following 
$$I(A) = \int d\theta_n\dots d\theta_1\,\exp\Bigg[\frac{1}{2}(\theta,A\theta)\Bigg]\\ = 
\int d\theta_n\dots d\theta_1\, (\theta_1\theta_2+\theta_3\theta_4+\cdots)\\=0$$
The answer should be $$I(A) = 1.$$ 
In the above I use (perhaps incorrectly?) 
$$\int d\theta_n\dots d\theta_1\, \theta_n\dots \theta_1\, = 1 $$
and
$$\int d\theta_n\dots d\theta_1= 0. $$
Where do I err? 
EDIT: I think I know how to fix this: it is the last term in the expansion of the exponential the contributes. All other terms give zero (just like the one above). I will add the solution later. 
 A: The problem is that you are generalizing the real Gaussian integral to the Grassmannian case rather than the complex Gaussian integral $\int dz d\bar{z} e^{-(z, Az)}$, where $(z, A z) = \sum_{i,j} z_i A_{ij} \bar{z}_j$ and $d z d\bar{z} = dz_1 d\bar{z}_1 \ldots dz_n d\bar{z}_n$.
You can see that the determinant will not arise by looking at the $2$-dimensional case. That is, let $A$ be a $2 \times 2$ matrix with entries $A_{ij}$. Then  $(\theta, A \theta) = \theta_1 A_{12} \theta_2 + \theta_2 A_{21} \theta_1$. The anti-commutativity kills the $A_{11}$ and $A_{22}$ that you need in the determinant of $A$.
To fix this, introduce "conjugate variables" $\bar{\theta}_1$ and $\bar{\theta}_2$. The bar doesn't denote any kind of operation on the $\theta_i$. The $\bar{\theta}_i$ are just new Grassmann variables. Then define
\begin{equation}
(\theta, A \bar{\theta}) = \theta_1 A_{11} \bar{\theta_1} + \theta_1 A_{12} \bar{\theta}_2 + \theta_2 A_{21} \bar{\theta}_1 + \theta_2 A_{22} \bar{\theta}_2.
\end{equation}
The terms $\theta_i A_{ii} \bar{\theta}_i$ no longer vanish because there are no relations between $\theta_i$ and $\bar{\theta}_i$. Now
\begin{equation}
\int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 e^{-(\theta, A \bar{\theta})} = \int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 \frac{1}{2} (\theta, A \bar{\theta})^2
\end{equation}
since the first two terms $1$ and $-(\theta, A \bar{\theta})$ in the exponential vanish under integration and the higher-order terms $(\theta, A \bar{\theta})^n$ for $n \geq 3$ vanish by anti-commutativity.
Now a computation shows that
$(\theta, A \bar{\theta})^2 = 2 \theta_1 \bar{\theta}_1 \theta_2 \bar{\theta}_2 \det A$,
so
\begin{equation}
\int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 e^{-(\theta, A \bar{\theta})}  = \det A.
\end{equation}
Given Grassmann variables $\theta_i, \bar{\theta}_i$ for $i = 1, \ldots, n$, defining $(\theta, A \bar{\theta}) = \sum_{i,j=1}^n \theta_i A_{ij} \bar{\theta}_j$ yields $\int d\theta_1 d\bar{\theta}_1 \ldots d\theta_n d\bar{\theta}_n e^{-(\theta, A \bar{\theta})} = \det A$ by a similar computation (this is really an exercise in bookkeeping).
A: You are absolutely right in your edit. Only the highest order term of the expansion survives integration. This is because $\int d\theta_i=0$ so that we need a term where all grassmann variables are there ($\int d\theta_n...d\theta_1 \theta_1...\theta_n=1$).
I noticed that this post is really old but still felt like replying as this connects to subjects maters I am currently studying about majorana fermions in physics.
In general one can always write a "gaussian" integral over real grassmann variables as
\begin{equation}
    \int d\theta_1d\theta_2...d\theta_N \exp\left\{-\frac{1}{2}\sum_{i,j=1}^N\theta_iA_{ij}\theta\right\}.
\end{equation}
Where $A$ is a $N\times N$ real skew-symmetric matrix. Now notice that such matrices have the following properties:
Assume $\vec{x}$ to be an eigenvector of $A$ than $A\vec{x}=\lambda\vec{x}$ and $(A\vec{x})^\dagger=(\lambda\vec{x})^\dagger \rightarrow \vec{x}^\dagger A^T=\lambda^*\vec{x}^\dagger$ using this
\begin{equation}
\label{Im}
\begin{split}
    \vec{x}^\dagger A^T\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\
    \vec{x}^\dagger (-A)\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\
    -\lambda\vec{x}^\dagger\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\
    -\lambda&=\lambda^*
\end{split}
\end{equation}
so that
\begin{equation}
\label{pairs}
\begin{split}
    (A\vec{x})^*&=(\lambda\vec{x})^*\\
    A\vec{x}^*&=\lambda^*\vec{x}^*\\
    A\vec{x}^*&=-\lambda\vec{x}^*.
\end{split}
\end{equation}
We have found that $\lambda_i$ is pure imaginary and that all eigenvalues appear in $\pm\lambda$ pairs. Consequently $A$ has the following spectrum of eigenvalues 
\begin{equation}
\label{spec}
    \lambda(A)=\{\pm i\lambda_1, \pm i\lambda_2,...,\pm i\lambda_m\}\quad\quad\text{Where}\quad\lambda_i\in\mathbb{R}
\end{equation}
As a side note observe that if $N$ is uneven that at least one of the eigenvalues must be zero. This is not the case for even $N$ where we have defined $(N=2m)$. Using the property that the determinant is so called similarity invariant (similar matrices give the same determinant) we can write
\begin{equation}
\label{det}
    \det(A)=\prod_{i=1}^m\lambda_i^2
\end{equation}
Also note that such a matrix can be put in to block-diagonal form by special-orthogonal transformation $D=OAO^T$ (see wiki) to get
\begin{equation}
\begin{split}
    D&=diag\left(\left(
    \begin{matrix}
        0 & \lambda_1 \\
        -\lambda_1 & 0
    \end{matrix}
    \right)
    ,
    \left(
    \begin{matrix}
        0 & \lambda_2 \\
        -\lambda_2 & 0
    \end{matrix}
    \right)
    ,...,
    \left(
    \begin{matrix}
        0 & \lambda_m \\
        -\lambda_m & 0
    \end{matrix}
    \right)\right)
\end{split}    
\end{equation}
We can now start looking at the integral. First lets show that the measure will be invariant under change of variables $O_{ij}\psi_j=\psi_i'$ since
\begin{equation}
\begin{split}
    \prod_i^N\theta_i'&=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}\theta_{i_1}'\theta_{i_2}'...\theta_{i_N}'\\
    &=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}O_{i_1,i_1'}\theta_{i_1'}O_{i_2,i_2'}\theta_{i_2'}...O_{i_N,i_N'}\theta_{i_N'}\\
    &=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}U_{i_1,i_1'}O_{i_2,i_2'}...O_{i_N,i_N'}\epsilon_{i_1',i_2',...,i_N'}\prod_i^N\theta_{i}\\
    &=\det(O)\prod_i^N\theta_{i}\\
    &=\prod_i^N\theta_{i}
\end{split}
\end{equation}
This means that (omitting the apostrophe's) we can write the integral in the following form and start to solve it
\begin{equation}
\begin{split}
    &\int d\theta_1d\theta_2...d\theta_N e^{-\frac{1}{2}\theta_iD_{ij}\theta_j}\\
    &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{-\frac{1}{2}\left(\lambda_1\theta_1\theta_2-\lambda_1\theta_2\theta_1+\lambda_2\theta_3\theta_4-\lambda_2\theta_4\theta_3+...+\lambda_m\theta_{m-1}\theta_m-\lambda_m\theta_{2m}\theta_{2m-1}\right)\right\}\\
    &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{\lambda_1\theta_2\theta_1+\lambda_2\theta_4\theta_3+...+\lambda_m\theta_{2m}\theta_{2m-1}\right\}\\
    &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{\sum_{i=1}^m\lambda_i\theta_{2i}\theta_{2i-1}\right\}\\
    &=\int d\theta_1d\theta_2...d\theta_N\prod_{i=1}^m\exp\left\{\lambda_i\theta_{2i}\theta_{2i-1}\right\}\\
    &=\int d\theta_1d\theta_2...d\theta_N\prod_{i=1}^m(1+\lambda_i\theta_{2i}\theta_{2i-1})\\
    &=\int d\theta_1d\theta_2...d\theta_N\left(\prod_{i=1}^m\lambda_i\right)\theta_2\theta_1\theta_4\theta_3...\theta_N\theta_{N-1}\\
    &=\prod_{i=1}^m\lambda_i\\
    &=\sqrt{\det(A)}
\end{split}
\end{equation}
Where we have used that the only the part of the product $(\prod_{i=1}^m(1+\lambda_i\theta_{i+1}\theta_i))$ that survives integration is where we multiply all the grassmann variables (All other terms will vanish because $\int d\theta=0$). At the end of the calculation I used that pairs of grassmann variables commute ([$\theta_2\theta_1,\theta_4\theta_3]=0$)
