Let $n\ge2$ find the sum of determinants where the values of $\alpha$ are given independently from one another from 1 to n Let $n\ge2$ find the sum of determinants where the values of $\alpha$ are given independently from one another from $1$ to $n$
 $\sum_{\alpha_{1},\alpha_{2},...,\alpha_{n},=1}^n$$
    \begin{vmatrix}
    a_{1\alpha_1} & a_{1\alpha_2} &\cdots& a_{1\alpha_n} \\
     a_{2\alpha_1} & a_{2\alpha_2} &\cdots& a_{2\alpha_n} \\
\vdots & \vdots &\ddots &\vdots\\
    a_{n\alpha_1} & a_{n\alpha_2} &\cdots& a_{n\alpha_n} \\
    \end{vmatrix}
$
Can someone help me with this problem? I tried solving it with induction but I don't know how to prove the $n+1$ step. I think the sum is $0$. I would appreciate some help
 A: Let $\left(\alpha_1,\dots,\alpha_n\right)$ be such that $1\leqslant \alpha_i\leqslant n$. If there exists $i,j\in\left\{1,\dots,n\right\}$ such that $i\neq j$ and $\alpha_i=\alpha_j$, then the corresponding determinant is zero because two columns are equal. Therefore, we have to restrict to the $\left(\alpha_1,\dots,\alpha_n\right)$ such that $\left\{\alpha_1,\dots,\alpha_n\right\}=\left\{1,\dots,n\right\}$. Therefore, the sum can be indexed by permutations and is equal to 
$$
S_n:=\sum_{\sigma\in\mathcal S_n}\det\left(A_\sigma\right)
$$
where $A_\sigma$ is the matrix with entries $\left(a_{i,\sigma(j)}\right)_{1\leqslant i,j\leqslant n}$. After having switched the columns, we derive that 
$$
S_n=\sum_{\sigma\in\mathcal S_n}\varepsilon\left(\sigma\right)  \det\left(A\right)
$$
where $\varepsilon\left(\sigma\right)$ denotes the signature of the permutation $\sigma$ and $A$ is the matrix with entries $\left(a_{i,j}\right)_{1\leqslant i,j\leqslant n}$. the sum of the signatures is indeed zero.
