Differential forms dimensions

Question

So I am studying differential geometry and I am struggling to understand how 1.18 comes from 1.17? If $p=1$ it must be $n$.

Answer 1

We have $$ \Lambda V = \bigoplus_{p=0}^\infty \Lambda^p V \implies \dim \Lambda V = \sum_{p=0}^\infty \dim \Lambda^p V $$ but you know that $$ \dim \Lambda^p V = 0\quad \text{for} \quad p>n $$ It follows from $(1.17)$ that $$ \dim \Lambda V = \sum_{p=0}^n \binom{n}{p} = 2^n \tag{1.18} $$