Let $\Omega^k (U)$ denote the set of differential $k$-forms on an open subset $U\subseteq \mathbb{R}^n$. For each $k\in \mathbb{N}$ the exterior derivative $d_k=d : \Omega^{k-1} (U) \rightarrow \Omega^k (U)$ sends the $(k-1)$-form $\omega$ to the $k$-form $d \omega$ defined by $d(\omega) = d(\omega_{i_1 \dots i_{k-1}} f^{(i_1)} \wedge \dots \wedge f^{(i_{k-1})})=d\omega_{i_1 \dots i_{k-1}}\wedge f^{(i_1)} \wedge \dots \wedge f^{(i_{k-1})}$ where we define the $1$-form $d\omega_{i_1 \dots i_{k-1}}=\frac{\partial}{\partial x^j}(\omega_{i_1 \dots i_{k-1}}) f^{(j)}$ (and $f^{(i)}$ are the dual basis vectors for ${\mathbb{R}^n}^*$). Since $d^2 = 0$, we obtain a sequence of vector spaces
$C^\infty (U) =\Omega^0 (U) \xrightarrow{d_1} \Omega^1 (U) \xrightarrow{d_2} \cdots\xrightarrow{d_k} \Omega^k (U) \xrightarrow{d_{k+1}}\cdots$
where $\operatorname{im} d_k \subseteq \ker d_{k+1}$ for all $k$. I haven't studied any (co)homology formally (although I have some limited experience now with modules and exact sequences) but I do recognise that this looks like a complex on which we might "do" cohomology. I asked my lecturer about this but I don't think he wanted to go into anything deeply - the differential geometry course I'm taking seems to be more on the concrete side with particular emphasis on subsets of $\mathbb{R}^n$ rather than general manifolds.
However I am really interested in the abstract side of things and I'd like to know more about this sequence. Could anyone tell me anything "interesting" about the above complex: for example, how can we go about studying the cohomology and can we use it to say anything useful about our spaces of forms? What objects of study does this complex lead us to? My question is intentionally vague because I'd like to see a range of answers so as to get a feel for the first real instance of cohomology that I've seen in my degree. Many thanks!
It is called De Rham complex and the cohomology it induces is called De Rham cohomology. $U$ can be replaced by any smooth manifold. It is naturally isomorphic to other cohomology theories with $\mathbb{R}$-coefficients; the exterior product of forms corresponds to the cup product in singular/simplicial cohomology.
Note also that for $U\subseteq \mathbb{R}^3$ open, this sequence can be easilly identified with the three operators gradient, curl and divergence, so De Rham generalizes the formulas $\text{curl} \,\,\text{grad}=\text{div}\,\, \text{curl}=0$. The first De Rham cohomology measures the obstruction, for a "field with curl zero" (if you identify a $1$-form with a vector field via some metric), to be a gradient of some scalar function, for example.