Operations in the exterior algebra. Multiplication in the direct sum of rings. Let $V$ be a vector space with dimension $n$ over a field $K$. The exterior algebra $\Lambda(V)$ of the vector space $V$ is the direct sum of the exterior powers $\Lambda^k(V),\quad k\in\overline{0,n}$. Then an element $x\in\Lambda(V)$ has the form $(x_0,\dots,x_n)$, where $x_i \in \Lambda^i(V)$ is the $i$-th homogeneous component of $x$. As each exterior power is itself a vector space over $K$, so is their direct sum with the point-wise addition and multiplication with a scalar.
How does the exterior product carry over the direct sum? What does $(x_0,\dots,x_n)\wedge (y_0,\dots,y_n)$ stand for?

The product is defined component-wise, i.e. $$(x_0,\dots,x_n)\wedge (y_0,\dots,y_n) = (z_{0}, \ldots z_{n})$$ where $$z_{i} = \sum_{j = 0}^{i} x_{j} \wedge y_{i-j}$$
