Why is Grassman integration so weird? Why are Grassman integration and differentiation equivalent? The only justification of this definition I have ever scene is "Well, how else could it work?" Indeed, I don't have any other suggestions, but I'd like something a bit more rigorous or formal, or at least more detailed. 
In particular, I am interested in this as it applies to fermionic systems in physics, since that is the primary context that I am familiar with. Explanations from different perspectives are welcome as well.
 A: One way to define Grassmann integration is via axioms. See for instance this answer on Physics.SE.
An equivalent way of thinking about it that I find very natural is the following: Grassmann integration is the integration of differential forms.
To see this, suppose you're working with $n$ Grassmann variables $\theta_1, \ldots, \theta_n$. That is, the $\theta_i$ are generators of an abstract Grassmann algebra, i.e. the polynomials in $\theta_i$ modulo the anti-commutativity relation that $\theta_i^2 = 0$ for each $i$.
If you're familiar with differential forms, it's easier to recast the above in that language. That is, just rename $\theta_i$ to $dx_i$ and think of the "abstract" Grassmann algebra above as the "concrete" exterior algebra (the algebra of differential forms) over $\mathbb{R}^n$. Then the anti-commutativity relation simply means that we multiply the forms via the wedge product (as usual). However, let's suppress that in the notation so that $dx_i dx_j$ means $dx_i \wedge dx_j$.
By anti-commutativity, we need only concern ourselves with polynomials $f(dx_1, \ldots, dx_n)$ in the $dx_i$ (analytic functions of forms can be defined by Taylor expansion, which necessarily terminates after a finite number of terms), which themselves are differential forms.
Now the Grassmann integral $\int f(\theta_1, \ldots, \theta_n) d\theta_1 \ldots d\theta_n$ is simply the integral $\int_{\mathbb{R}^n} f(dx_1, \ldots, dx_n)$ of the form $f(dx_1, \ldots, dx_n)$ over $\mathbb{R}^n$ (I didn't include any "extra" differentials in the second integral because the integrand is already a differential form). What's crucial is that we are (essentially) only integrating the top-degree part of this form (because the integral of a $k$-form over a space of dimension $n > k$ is $0$ by definition). This reflects the fact that, e.g. $\int \theta_1 d\theta_1 d\theta_2 = 0$ in Grassmann integration.
It's a good exercise to see that integrating forms in this way respects the axioms of Grassmann integration.
