I am trying to prove the following:
$$ \det\begin{pmatrix} \alpha v_1\\ v_2\\ \vdots\\ v_n \end{pmatrix}=\alpha \cdot\det\begin{pmatrix} v_1\\ v_2\\ \vdots\\ v_n \end{pmatrix}$$
Let $B$ be the matrix $A$ with a row multiplied by $\alpha$, wlog let call this row $r^j$. For each $\sigma\in S_n$ due to the fact that a permutation is a bijection there will be an element of the row $r^j$ so we will get
$$\det(B)=\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^{n}b_{(i,\sigma(i))}=\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\cdot\alpha a_{(j,\sigma(j))}\prod_{i\ne j} a_{(i,\sigma(i))}$$ hence $$\det(B)=\alpha \cdot \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\cdot a_{(j,\sigma(j))}\prod_{i\ne j} a_{(i,\sigma(i))}=\alpha \cdot \det(A)$$ where $a_{(j,\sigma(j))}\in r^j$.