Bott and Tu - Page 13
Let $x_1,\dots,x_n$ be the coordinates of $\Bbb R^n$. We define $\Omega^*$ to be the algebra over $\Bbb R$ generated by $dx_1,\dots,dx_n$ subject to the relations: $$\begin{cases}(dx_i)^2=0\\ dx_idx_j=-dx_jdx_i,\quad i\ne j\end{cases}$$
Why exactly do we need to state the both of these. The first would imply the second if we had bilinearity (meaning I relate this to the exterior algebra $\bigwedge \Bbb R^n$). But the latter condition, in any case, implies the former, so long as we lift the requirement that $i\ne j$.
Indeed, $dx_idx_i=-dx_idx_i\iff 2dx_i^2=0$ and evidently, $2\in \Bbb R$ is not a zero divisor, and hence $dx_i^2=0$ as in the first relation.
Even if we were defining this for a field of characteristic $2$, the latter condition would immediately imply the former.
Why write both relations? Why not lift $i\ne j$ in the second?
You're completely right. It's just for emphasis (and to help the reader get the important points). In particular, it might make it slightly clearer that $dx_idx_j$ with $i<j$ gives a basis for the vector space $\bigwedge^2\Bbb R^2$.
By the way, I love Bott/Tu, but the fact that they omit the wedge products throughout will be confusing to a reader who goes on to study something like Riemannian geometry, where one writes $dx_i^2$ for $dx_i\otimes dx_i$ (in the context of Riemannian metrics). So I really prefer to use the wedge products, even though it's a bit more writing/typing/printing.