This is pretty much a linear algebra question.
Let $\omega$ be an alternating $k$-form on vector space $\mathbb{R}^n$ and $k>n$. Then because of linearity, for $k$ vectors $\vec{v_k} = \sum_{i=1}^n \alpha_{i,k} \vec{e_i}$ (where $\vec{e_i}$ are basis vectors), $\omega(v_1,\dots,v_k)$ is a sum of terms of the form $\alpha \omega(\vec{e_{i_1}}, \dots, \vec{e_{i_k}})$ where $\alpha \in \mathbb{R}$ is a scalar. Since $k>n$, we can always find indexes $p$ and $q$, $p \neq q$, such that $\vec{e_{i_p}} = \vec{e_{i_q}}$ and because of that $\omega = 0$.
For the second part of the question, if it were $a_2 = \alpha a_1$ then $a_1 \wedge a_2 \wedge \dots \wedge a_p = \alpha a_1 \wedge a_1 \wedge \dots \wedge a_p = 0$.
On the other hand, if $1$-forms $a_1, a_2, \dots, a_p$ were linearly independent, we could extend this set to a basis $a_1, \dots, a_p, a_{p+1}, \dots, a_n$ of the dual space $(\mathbb{R}^{n})^*$. It is known that the space $\Lambda^n \mathbb{R}^n$ of alternating $n$-forms is one-dimensional. Suppose $a_1 \wedge \dots \wedge a_n = 0$. Let $\omega = b_1 \wedge \dots \wedge b_n$ be an alternating $n$-form and $b_i = A a_i$. Then $\omega = A a_1 \wedge \dots \wedge A a_n = \det(A) a_1 \wedge \dots \wedge a_n = 0$. It follows that every alternating $n$-linear form is equal to $0$ which is a contradiction.
"Finite-dimensional Vector Spaces" by Halmös has a nicely written chapter about multilinear forms.